\u4e0d\u77e5\u4e3a\u5565\u6ca1\u4e86\u5c45\u7136\u2026\u2026\u2026\u2026\u611f\u8c22google\u5feb\u7167\uff0c\u6015\u518d\u51fa\u4e8b\u6211\u8fd8\u662f\u8f6c\u5230\u8fd9\u4e00\u4e0b\u5427\u2026\u2026\u2026\u2026<\/p>\n
<\/p>\n
\u672c\u6765\u51c6\u59078\u9898\u7684\uff0c\u6700\u540e\u4e00\u9898\u60f3\u51fa\u4e00\u4e2a\u8ba1\u7b97\u6bd4\u8f83\u7e41\u7410\u7b97\u5b8c\u5c31\u76f4\u63a5\u5f88\u7b80\u5355\u7684coding\u7684\u6570\u5b66\u9898\uff0c\u4f46\u60f3\u6765\u60f3\u53bb\u51fa\u4e0d\u6765\u2026\u2026\u4e8e\u662f\u5c317\u9898\u4e86\u3002<\/p>\n
\u7279\u522b\u6c34\u7684\u9898\u6bd4\u8f83\u5c11\uff0c\u4f46\u6211\u89c9\u5f97\u5982\u679c\u7b2c\u4e00\u9898\u77e5\u9053long long\u4e4b\u540e\uff0c\u51fa\u4e24\u9898\u4e0d\u5e94\u8be5\u662f\u95ee\u9898\uff0c\u4f46\u597d\u50cf\u8fd9\u6837\u7684\u5927\u4e00\u65b0\u751f\u4e5f\u4e0d\u7b97\u591a\u3002\u4ece\u6392\u540d\u4e0a\u770b\uff0c\u7b2c\u4e00\u7684\u5362\u5927\u795e\u662f\u54c8\u5de5\u5927\u6709\u60c5\u53c2\u8d5b\u7684\uff0c\u8ba9\u5927\u5bb6\u770b\u770b\u4e9a\u6d32\u8d5b\u94f6\u724c\u7684\u5b9e\u529b\u3002\u4e1c\u5927\u7b2c\u4e00\u540d\u662f\u552f\u4e00\u4e00\u4e2a\u53c2\u52a0\u4eca\u5e74regional\u768410\u7ea7\u7f57\u8000\u71ca\uff0c\u53ef\u4ee5\u8bf4\u5c31\u662f\u73b0\u5f79\u9009\u624b\u4e2d\u660e\u5e74\u6700\u6709\u524d\u9014\u7684\u4e86\u3002\u65b0\u624b\u7b2c\u4e00\u540d\u662f\u603b\u7b2c\u4e09\u768410\u7ea7\u9093\u6500\u5cf0\uff0c\u4eba\u5bb6\u53ef\u662f\u59b9\u5b50\uff0c\u6bd4\u5927\u591a\u6570\u7537\u4eba\u5f3a\u591a\u4e86\u300211\u7ea7\u201c\u65b0\u661f\u201d\u7b2c\u4e00\u540d\u662f\u603b\u7b2c\u516d\u7684\u674e\u91d1\u9e4f\uff0c\u80fd\u8fc7G\u9898\u8fd8\u662f\u975e\u5e38\u5389\u5bb3\u7684\uff0c\u524d\u9014\u65e0\u91cf~<\/p>\n
\u8be6\u7ec6\u6700\u7ec8\u6392\u540d\u548c\u79ef\u5206\u5c06\u5728\u7a0d\u540e\u53d1\u5e03\u3002<\/p>\n
A\u9898\u5c31\u662flong long\u7684\u95ee\u9898\u3002\u5218\u8001\u5e08\u8bf4\u4ed6\u7b2c\u4e00\u5468\u5c31\u7559\u4f5c\u4e1a\u8ba9\u5927\u5bb6\u67e5C\/C++\u6807\u51c6\u6570\u636e\u7c7b\u578b\uff0c\u8f6f\u4ef6\u7684\u57f9\u8bad\u4e5f\u63d0\u5230\u8fc7long long\uff0c\u786e\u5b9e\u4e0d\u5e94\u8be5\u8fd9\u4e48\u591a\u4eba\u4e0d\u77e5\u9053\u3002\u540e\u6765\u6211\u6015\u5927\u5bb6\u90fd0\u9898\u5c31\u76f4\u63a5\u544a\u8bc9\u5404\u4f4d\u600e\u4e48\u5199\u4e86\uff0c\u6240\u4ee5\u5bfc\u81f4\u5f88\u591a\u4eba\u90fd\u662f1\u5c0f\u65f640\u5206\u8fc7\u7684\uff0c\u8fd9\u6837\u8c01\u4e4b\u524d\u6ca1\u63d0\u4ea4\u8c01\u5c31\u9760\u524d\uff0c\u6240\u4ee5\u8fd9\u9898\u7684\u6392\u540d\u6709\u70b9\u6ca1\u542b\u91d1\u91cf\u3002\u4e0d\u8fc7\u603b\u4f53\u4e0a\u5f71\u54cd\u4e0d\u5927\u3002<\/p>\n
B\u9898\u4e5f\u5f88\u6c34\uff0c\u8bfb\u5165\u5224\u65ad\u662f\u5426\u662f11\u7ea7\uff0c\u518d\u5224\u65ad\u662f\u5426\u91cd\u590d\u548c\u6027\u522b\uff0c\u8f93\u51fa\u5373\u53ef\u3002\u6ca1\u505a\u51fa\u6765\u7684\u518d\u4ed4\u7ec6\u770b\u4e0b\u4ee3\u7801\u6709\u4ec0\u4e48\u95ee\u9898\u3002<\/p>\n
\u524d\u4e24\u9898\u5c5e\u4e8e\u5fc5\u4f1a\u4e5f\u662f\u57fa\u672c\u6240\u670911\u7ea7\u53ea\u4f1a\u7684\u4e24\u9898\uff0c\u786e\u5b9e\u6574\u4f53\u96be\u5ea6\u5bf9\u4e8e\u65b0\u624b\u5927\u4e86\u70b9\uff0c\u8fd9\u4e2a\u602a\u6211\u4e86\u3002\u3002\u3002\u56e0\u4e3a\u8fc7\u4e8e\u57fa\u7840\u6240\u4ee5\u4e0d\u53d1\u4ee3\u7801\u4e86\uff0c\u81ea\u5df1\u5b9e\u73b0\u3002<\/p>\n
C\u9898\u5c5e\u4e8e\u6a21\u62df\u9898\uff0c\u601d\u8def\u5f88\u6e05\u6670\uff0c\u800c\u4e14\u56e0\u4e3a\u90fd\u662f\u4eba\u8138\u7684\u7167\u7247\u663e\u7136\u4e0d\u5b58\u5728\u4fe9\u773c\u775b\u5728\u9f3b\u5b50\u4e00\u5934\u7684\u60c5\u51b5\uff0c\u5148\u628a\u8138\u7684\u8f6e\u5ed3\u5224\u65ad\u51fa\u6765\u518d\u5206\u522b\u5224\u65ad\u5404\u4e2a\u5668\u5b98\u7684\u6761\u4ef6\u5c31\u53ef\u4ee5\u4e86\u3002\u6bd4\u8d5b\u7684\u65f6\u5019\u5c45\u7136\u53d1\u73b0\u6709\u4eba\u4f01\u56fe\u5168\u8f93\u51faNo\u9a97\u5206\uff0c\u6570\u636e\u663e\u7136\u4e0d\u4f1a\u8fd9\u4e48\u5f31\u2026\u2026
\n\u53d1\u4e0b\u5362\u5927\u795e\u7684\u4ee3\u7801\u3002
\n#include <iostream>
\n#include <cstdio>
\n#include <cstring>
\n#include <algorithm>
\n#include <cmath>
\n#include <vector>
\nusing namespace std;
\nconst int step[][2] = {\u00a0\u00a0{0,1} , {0,-1} ,{1, 0}, {-1,0} };
\nchar mat[21][41];
\nbool chk[21][41];
\nchar s1[41],s2[41],buf[41];
\nint Left[21],Right[21],h[41];
\nvoid dfs(int i,int j,int d)
\n{
\nif(chk[i][j]) return;
\nchk[i][j] = true;
\nmat[i][j] = ‘0’;
\nfor(int k = 0 ; k < 4 ; k++)
\n{
\nint ty = i + step[k][0];
\nint tx = j + step[k][1];
\nif(ty>=0 && ty < 20 && tx>=0 && tx < 40)
\n{
\nif(d == 1)
\n{
\nif(mat[ty][tx] == ‘1’)
\ndfs(ty,tx,d);
\n}
\nelse
\n{
\nif(mat[ty][tx] == ‘0’)
\ndfs(ty,tx,d);
\nelse
\ndfs(ty,tx,1);
\n}
\n}
\n}<\/p>\n
}
\nvoid debug()
\n{
\nfor(int i = 0; i< 20 ; i ++)
\nprintf(“%sn”,mat[i]);
\n}
\nvoid dfs(int i ,int j,vector<pair<int,int> >& v)
\n{
\nif(chk[i][j]) return;
\nchk[i][j] = true;
\nmat[i][j] = ‘ ‘;
\nv.push_back(make_pair(i ,j));
\nfor(int k = 0 ; k < 4 ; k++)
\n{
\nint ty = i + step[k][0];
\nint tx = j + step[k][1];
\nif(ty>=0 && ty < 20 && tx>=0 && tx < 40)
\n{
\nif(mat[ty][tx] == ‘1’)
\ndfs(ty,tx,v);
\n}
\n}
\n}
\nvoid gao(int i,int j)
\n{
\nif(chk[i][j]) return;
\nchk[i][j] = true;
\nmat[i][j] =’ ‘;
\nLeft[i] = min(Left[i],j);
\nRight[i] = max(Right[i],j);
\nfor(int k = 0 ; k < 4 ; k++)
\n{
\nint ty = i + step[k][0];
\nint tx = j + step[k][1];
\nif(ty>=0 && ty < 20 && tx>=0 && tx < 40)
\n{
\nif(mat[ty][tx] == ‘1’)
\ngao(ty,tx);
\n}
\n}
\n}<\/p>\n
void checkeyebrow(bool& ok,int& line)
\n{
\nfor(; line < 20 ; line ++)
\n{
\nfor(int i = 0 ; i < 40 ; i ++)
\nbuf[40 – i – 1] = mat[line][i];
\nbuf[40] = 0;
\nif(sscanf(mat[line],”%s %s”,s1,s2) == 2)
\n{
\nsscanf(mat[line],”%s”,s1);
\nsscanf(buf,”%s”,s2);
\nif(strcmp(s1,s2))
\nok = false;
\nfor(int i = 0 ; i < 40 ; i++)
\nmat[line][i] = ‘ ‘;
\nbreak;
\n}
\n}
\n}
\nvoid checkeye(bool& ok,int& line)
\n{
\nfor(; line < 20 ; line ++)
\n{
\nif(sscanf(mat[line],”%s %s”,s1,s2) == 2)
\n{
\nvector< pair<int,int> > v1,v2;
\nfor(int i = 0 ; i < 40 ; i ++)
\n{
\nif(mat[line][i] == ‘1’)
\n{
\nmemset(chk,0,sizeof(chk));
\ndfs(line,i,v1);
\nbreak;
\n}
\n}
\nfor(int i = 40 – 1 ; i >=0 ; i –)
\n{
\nif(mat[line][i] == ‘1’)
\n{
\nmemset(chk,0,sizeof(chk));
\ndfs(line,i,v2);
\nbreak;
\n}
\n}
\nsort(v1.begin() , v1.end());
\nsort(v2.begin() , v2.end());
\nif(v1.size()!= v2.size())
\n{
\nok = false;
\nreturn;
\n}
\nfor(int i = 1; i < v1.size() ; i ++)
\n{
\nif( (v1[i].first – v1[0].first != v2[i].first – v2[0].first) || (v1[i].second – v1[0].second\u00a0\u00a0!= v2[i].second\u00a0\u00a0– v2[0].second ))
\n{
\nok = false;
\nbreak;
\n}
\n}
\nreturn;
\n}
\n}
\n}
\nvoid checknose(bool& ok,int& line)
\n{
\nfor(; line < 20 ; line ++)
\n{
\nif(sscanf(mat[line],”%s”,s1) == 1)
\n{
\nfor(int i = 0 ; i < 40 ; i++)
\n{
\nif(mat[line][i] == ‘1’)
\n{
\nmemset(chk,0,sizeof(chk));
\nmemset(Right,-1,sizeof(Right));
\nmemset(Left,0x3f,sizeof(Left));
\ngao(line ,i);
\nfor(int j = line + 1; j < 20; j ++)
\n{
\nif(Right[j] == -1)
\nbreak;
\nif( (Left[j] + Right[j]) != (Left[line] + Right[line]))
\n{
\nok = false;
\nbreak;
\n}
\n}
\nreturn;
\n}
\n}
\n}
\n}
\n}
\nvoid checkmouth(bool& ok,int& line)
\n{
\nmemset(h,-1,sizeof(h));
\nfor(int i = 0 ; i < 20 ; i ++)
\nfor(int j = 0; j < 40 ; j++)
\nif(mat[i][j] == ‘1’)
\n{
\nmat[i][j] =’ ‘;
\nif(h[j] == -1)
\nh[j] = i;
\nelse
\nh[j] = min(h[i],i);
\n}
\nint i = 0;
\nwhile(h[i] == -1)
\ni++;
\nint j = i;
\nwhile(h[j] != -1)
\nj++;
\nj–;
\nif( (j – i) % 2 == 0)
\n{
\nint mid = (i + j) \/ 2;
\nif(h[i] >= h[mid] || h[j] >= mid)
\n{
\nok = false;
\n}
\n}
\nelse
\n{
\nint mid = (i + j) \/ 2;
\nif(h[i] >= h[mid] || h[j] >= mid)
\n{
\nok = false;
\n}
\nmid = (i + j) \/ 2 + 1;
\nif(h[i] >= h[mid] || h[j] >= mid)
\n{
\nok = false;
\n}
\n}
\nreturn;
\n}
\nint main()
\n{
\nint t;
\nscanf(“%d”,&t);
\nwhile(t–)
\n{
\nfor(int i = 0; i< 20; i ++)
\nscanf(“%s”,mat[i]);
\nmemset(chk,0,sizeof(chk));
\ndfs(0,0 , 0);
\nfor(int i = 0 ; i < 20 ; i ++)
\nfor(int j = 0 ; j < 40 ; j ++)
\nif(mat[i][j] == ‘0’)
\nmat[i][j] = ‘ ‘;
\nint line = 0;
\nbool ok = true;<\/p>\n
checkeyebrow(ok,line);
\n\/\/debug();
\ncheckeye(ok,line);
\n\/\/debug();
\nchecknose(ok,line);
\n\/\/debug();
\ncheckmouth(ok,line);
\n\/\/debug();
\nok ? puts(“YES”) :puts(“No”);
\n}
\nreturn 0;
\n}<\/p>\n
\/**************************************************************
\nProblem: 1113
\nUser: luyi0619
\nLanguage: C++
\nResult: Accepted
\nTime:1 ms
\nMemory:1264 kb
\n****************************************************************\/
\n\u590d\u5236\u4ee3\u7801
\nD\u9898\u5c31\u662f\u4e2a\u7279\u522b\u88f8\u7684DP\u6709\u6728\u6709\u2026\u2026\u5927\u4e00\u4e0d\u4f1a\u7684\u6b63\u5e38\u4e3a\u5565\u6ca1\u4eba\u8fc7\u5462\u3002\u3001
\nN\u4e2a\u4eba\u5206\u6210M\u7ec4\uff0c\u5b50\u72b6\u6001\u80af\u5b9a\u5c31\u662f\u5c11\u4e8eN\u4e2a\u4eba\u5206\u6210M-1\u7ec4\uff0c\u51b3\u7b56\u662f\u5224\u65adM-1\u7ec4\u8ddf\u7b2cM\u7ec4\u5728\u54ea\u5206\u754c\u3002\u56e0\u6b64\u6700\u5916\u5c42\u4e24\u4e2a\u5faa\u73afN\u548cM\uff0c\u5148\u540e\u65e0\u6240\u8c13\u3002
\nf[n][m]=max{f[k][m-1]+mul(k+1,n)}
\nmul\u8868\u793a\u4ece\u7b2ck+1\u5230\u7b2cn\u4e2a\u7684\u4e58\u79ef\uff0c\u663e\u7136\u679a\u4e3ek\u4ece\u5927\u5f80\u5c0f\u5c31\u53ef\u4ee5\u76f4\u63a5\u5f97\u5230\u6bcf\u4e00\u6b65\u7684\u503c\u4e0d\u7528\u989d\u5916\u7b97\u4e86\uff0c\u56e0\u6b64\u6700\u540e\u590d\u6742\u5ea6\u662fn^2*m\u7684\u3002
\n\u6211\u7684\u4ee3\u7801\uff1a
\n#include<iostream>
\n#include<cstdio>
\n#include<cstring>
\nusing namespace std;<\/p>\n
long long a[100];
\nlong long f[100][15];<\/p>\n
int main()
\n{
\nint n,m;
\nwhile(scanf(“%d%d”,&n,&m)&&n>0)
\n{
\nmemset(f,0,sizeof(f));
\nfor(int i=1;i<=n;i++)
\n{
\nscanf(“%d”,&a[i]);
\n}
\nf[1][1]=a[1];
\nfor(int i=2;i<=n;i++)
\nf[i][1]=f[i-1][1]*a[i];
\nfor(int i=2;i<=n;i++)
\nfor(int j=2;j<=m;j++)
\n{
\nlong long mul=1;
\nfor(int k=i-1;k>=j-1;k–)
\n{
\nmul*=a[k+1];
\nf[i][j]=max(f[i][j],f[k][j-1]+mul);
\n}
\n}
\nprintf(“%lldn”,f[n][m]);
\n}<\/p>\n
return 0;
\n}
\n\/**************************************************************
\nProblem: 1114
\nUser: admin
\nLanguage: C++
\nResult: Accepted
\nTime:8 ms
\nMemory:1272 kb
\n****************************************************************\/
\n\u590d\u5236\u4ee3\u7801
\nE\u9898\u662f\u88f8\u6700\u5c0f\u751f\u6210\u6811\uff0c\u53ea\u8981\u4f1a\u7684\u90fd\u4e0d\u5e94\u8be5\u505a\u4e0d\u51fa\u6765\u3002OIer\u5982\u679c\u4e0d\u77e5\u9053\u600e\u4e48\u7528\u6a21\u7248\u81ea\u5df1\u5199\u4e5f\u4e0d\u5e94\u8be5\u662f\u95ee\u9898\uff0c\u5c31\u770b\u80fd\u4e0d\u80fd\u56de\u5fc6\u8d77\u6765\u6216\u8005\u6700\u8fd1\u6709\u6ca1\u6709\u590d\u4e60\u4e86\u2026\u2026\u6ce8\u610f\u8f93\u5165\u6570\u636e\u4e0d\u4e00\u5b9a\u975e\u662f\u6574\u6570\u3002
\n#include<cstdio>
\n#include<iostream>
\n#include<math.h>
\nusing namespace std;<\/p>\n
double px[1005],py[1005];
\ndouble mat[1005][1005];<\/p>\n
#define MAXN 1005<\/p>\n
#define inf 1000000000<\/p>\n
typedef double elem_t;
\nelem_t prim(int n,elem_t mat[][MAXN],int* pre){
\nelem_t min[MAXN],ret=0;
\nint v[MAXN],i,j,k;
\nfor (i=0;i<n;i++)
\nmin[i]=inf,v[i]=0,pre[i]=-1;
\nfor (min[j=0]=0;j<n;j++){
\nfor (k=-1,i=0;i<n;i++)
\nif (!v[i]&&(k==-1||min[i]<min[k]))
\nk=i;
\nfor (v[k]=1,ret+=min[k],i=0;i<n;i++)
\nif (!v[i]&&mat[k][i]<min[i])
\nmin[i]=mat[pre[i]=k][i];
\n}
\nreturn ret;
\n}<\/p>\n
int main()
\n{
\nint n;
\nwhile(scanf(“%d”,&n)>0)
\n{
\nfor(int i=0;i<n;i++)
\nscanf(“%lf%lf”,&px[i],&py[i]);
\nfor(int i=0;i<n;i++)
\nfor(int j=0;j<n;j++)
\n{
\nmat[i][j]=sqrt((px[i]-px[j])*(px[i]-px[j])+(py[i]-py[j])*(py[i]-py[j]));
\n}
\nint pre[MAXN];
\nprintf(“%.2fn”,prim(n,mat,pre));<\/p>\n
}<\/p>\n
return 0;
\n}
\n\/**************************************************************
\nProblem: 1115
\nUser: admin
\nLanguage: C++
\nResult: Accepted
\nTime:156 ms
\nMemory:9168 kb
\n****************************************************************\/
\n\u590d\u5236\u4ee3\u7801
\nF\u9898\u770b\u4e0a\u53bb\u7279\u522b\u590d\u6742\uff0c\u5b9e\u9645\u4e0a\u7279\u522b\u7b80\u5355\u3002\u770b\u4e0a\u53bb\u662f\u8ba1\u7b97\u51e0\u4f55\uff0c\u5b9e\u9645\u4e0a\u7b49\u4e8e\u6a21\u62df\u9898\u3002\u770b\u4e0a\u53bb\u4e0d\u77e5\u9053\u600e\u4e48\u627e\u70b9\u7684\u4f4d\u7f6e\u8ddf\u65b9\u5411\uff0c\u4f46\u5b9e\u9645\u4e0a\u663e\u7136\u4e34\u754c\u70b9\u5e94\u81f3\u5c11\u8ddf\u4e24\u4e2a\u591a\u8fb9\u5f62\u5904\u4e8e\u6070\u597d\u76f8\u5207\u7684\u72b6\u6001\uff0c\u6216\u8005\u8bf4\u5e94\u8be5\u81f3\u5c11\u8fc7\u5176\u4e2d\u4e24\u4e2a\u591a\u8fb9\u5f62\u7684\u67d0\u4e2a\u9876\u70b9\uff08\u60f3\u660e\u767d\u8fd9\u6b65\u5c31\u4e0d\u96be\u4e86\uff09\u3002\u4e8e\u662f\u76f4\u63a5\u679a\u4e3e\u4e24\u4e2a\u591a\u8fb9\u5f62\uff0c\u6bcf\u4e2a\u591a\u8fb9\u5f62\u679a\u4e3e\u4e00\u4e2a\u9876\u70b9\uff0c\u8fc7\u8fd9\u4fe9\u70b9\u7684\u76f4\u7ebf\u5148\u770b\u662f\u5426\u8fc7y\u8f74\u7279\u5b9a\u533a\u57df\uff0c\u7136\u540e\u679a\u4e3e\u5176\u4ed6\u591a\u8fb9\u5f62\u770b\u51e0\u4e2a\u76f8\u4ea4\u3002
\n\u7f57\u8000\u71ca\u7684\u4ee3\u7801\uff1a
\n#include<stdio.h>
\n#include<stdlib.h>
\n#include<math.h>
\n#define EX 10e-8
\ntypedef struct {double x,y;} Point;
\ntypedef struct {int p; Point point[100];} Polygon;
\nPolygon polygon[10000];
\ndouble y[10000];<\/p>\n
int cmp(const void* _a,const void* _b)
\n{
\nif ((double*)_a-(double*)_b>0) return 1;
\nelse return -1;
\n}<\/p>\n
double det(double x1,double y1,double x2,double y2)
\n{
\nreturn x1*y2-x2*y1;
\n}<\/p>\n
double cross(Point a,Point b,Point c)
\n{
\nreturn det(b.x-a.x,b.y-a.y,c.x-b.x,c.y-b.y);
\n}<\/p>\n
int dblcmp(Point a,Point b,Point c)
\n{
\nif (fabs(cross(a,b,c))<EX) return 0;
\nelse return cross(a,b,c)>0?1:-1;
\n}<\/p>\n
int intersect(Point a,Point b,double x,double y)
\n{
\nPoint c,d;
\nc.y=x; c.x=0;<\/p>\n
d.y=y; d.x=100;<\/p>\n
if (dblcmp(a,b,c)*dblcmp(a,b,d)<=0 && dblcmp(c,d,a)*dblcmp(c,d,b)<=0) return 1;
\nelse return 0;
\n}<\/p>\n
int check(Polygon p,double x,double y)
\n{
\nint i,flag=0;
\nPoint a,b;<\/p>\n
a = p.point[0];<\/p>\n
for(i=1;i<p.p;i++)
\n{
\nif (intersect(p.point[i],a,x,y)==1) {flag=1;
\nbreak;}
\na=p.point[i];
\n}<\/p>\n
return flag;
\n}<\/p>\n
int main()
\n{
\nint k,i,t,n,m,j,l,ans,max;
\nPoint p;<\/p>\n
scanf(“%d”,&t);<\/p>\n
for(i=1;i<=t;i++)
\n{
\nscanf(“%d”,&n);
\nmax=0;
\nm=0;<\/p>\n
for(j=0;j<n;j++)
\n{
\nscanf(“%d”,&polygon[j].p);<\/p>\n
for(l=0;l<polygon[j].p;l++)
\n{
\nscanf(“%lf%lf”,&polygon[j].point[l].x,&polygon[j].point[l].y);
\ny[m++]=polygon[j].point[l].y;
\n}
\n}<\/p>\n
for(j=0;j<m;j++)
\nfor(k=0;k<m;k++)
\n{
\nans=0;
\nfor(l=0;l<n;l++)
\n{
\nans+=check(polygon[l],y[j],y[k]);
\n}
\nif (ans>max) max=ans;<\/p>\n
}<\/p>\n
printf(“Case %d:%dn”,i,max);
\n}
\nreturn 0;
\n}<\/p>\n
\/**************************************************************
\nProblem: 1116
\nUser: Lyshen
\nLanguage: C++
\nResult: Accepted
\nTime:12 ms
\nMemory:16524 kb
\n****************************************************************\/
\n\u590d\u5236\u4ee3\u7801
\nG\u9898\u6ca1\u60f3\u5230\u8fd9\u4e48\u591a\u4eba\u5728\u4e0d\u4f1a\u5355\u8c03\u961f\u5217\u7684\u60c5\u51b5\u4e0b\u505a\u51fa\u4e86\u7c7b\u4f3c\u7684\u4f18\u5316\u3002\u6b64\u9898\u539f\u578b\u662fPOJ\u67d0\u4e2a\u9898\uff0c\u6211\u4eec\u6570\u636e\u5e94\u8be5\u66f4\u5f31\u3002\u9996\u5148\u7eaf\u679a\u4e3e\u5224\u65ad\u80af\u5b9a\u662fTLE\uff0c\u7136\u540e\u53ef\u4ee5\u770b\u5230\uff0c\u6bcf\u6b21\u53ea\u662f\u589e\u52a0\u4e00\u4e2a\u6570\uff0c\u5220\u9664\u4e00\u4e2a\u6570\uff0c\u90a3\u4e48\u5220\u9664\u4e4b\u540e\u6700\u5927\uff08\u6700\u5927\u6700\u5c0f\u4e00\u6837\uff0c\u6240\u4ee5\u7b97\u4e24\u6b21\u5c31\u53ef\u4ee5\uff0c\u5148\u6309\u7167\u6700\u5927\u8bb2\uff09\u5c31\u53ef\u80fd\u6539\u53d8\uff0c\u4f46\u6211\u4eec\u4e0d\u80fd\u53ea\u4fdd\u5b58\u6700\u5927\u662f\u591a\u5c11\uff0c\u56e0\u4e3a\u6700\u5927\u7684\u4e00\u65e6\u51fa\u4e86\u533a\u57df\u5c31\u4e0d\u77e5\u9053\u4e0b\u4e00\u5927\u7684\u6570\u8be5\u662f\u4ec0\u4e48\u3002\u56e0\u6b64\uff0c\u53ef\u4ee5\u7ef4\u62a4\u4e00\u4e2a\u961f\u5217\uff0c\u8fd9\u4e2a\u961f\u5217\u6309\u987a\u5e8f\u4fdd\u5b58\u6700\u5927\u5230\u6700\u5c0f\u7684\u6570\u5b57\u3002\u53ef\u4ee5\u770b\u51fa\uff0c\u5982\u679c\u540e\u63d2\u5165\u7684\u6bd4\u4e4b\u524d\u63d2\u5165\u7684\u6570\u8981\u5927\uff0c\u90a3\u4e48\u524d\u9762\u7684\u6570\u5c31\u6ca1\u5fc5\u8981\u7559\u7740\u56e0\u4e3a\u6c38\u8fdc\u4e0d\u4f1a\u53d8\u6210\u6700\u5927\u503c\u3002\u4e8e\u662f\u8fd9\u4e2a\u961f\u5217\u53eb\u201c\u5355\u8c03\u961f\u5217\u201d\uff0c\u6bcf\u6b21\u63d2\u5165\u4e4b\u524d\u90fd\u4fdd\u8bc1\u91cc\u9762\u662f\u5355\u8c03\u9012\u51cf\u7684\uff0c\u800c\u63d2\u5165\u4e4b\u524d\u5224\u65ad\u6700\u540e\u4e00\u4e2a\u5143\u7d20\u662f\u5426\u5c0f\u4e8e\u7b49\u4e8e\u8981\u63d2\u5165\u7684\u6570\uff0c\u5982\u679c\u662f\u5219\u5220\u9664\uff0c\u4e0d\u65ad\u5220\u9664\u76f4\u5230\u90fd\u5927\u4e8e\u6216\u8005\u961f\u5217\u4e3a\u7a7a\u3002\u8fd9\u6837\u6bcf\u6b21\u67e5\u8be2\u76f4\u63a5\u8fd4\u56de\u961f\u9996\uff0c\u6bcf\u6b21\u63d2\u5165\u4e4b\u524d\u5148\u5220\u9664\u9700\u8981\u5220\u9664\u7684\u5143\u7d20\uff0c\u518d\u63d2\u5165\u3002
\n\u6b64\u5916\u8fd9\u9898\u7ebf\u6bb5\u6811\u597d\u50cf\u53ef\u4ee5\uff0cRMQ\u80af\u5b9a\u53ef\u4ee5\u4f46\u4f1a\u6162\u4e00\u4e9b\u3002\u5176\u4ed6\u4eba\u60f3\u7684\u53ef\u80fd\u662f\u81ea\u5df1\u7684\u5404\u79cd\u4f18\u5316\u65b9\u6cd5\uff0c\u6709\u7684\u6bd4\u6211\u5199\u7684\u8fd8\u5feb\uff0c\u5982\u679c\u601d\u8def\u8ddf\u8fd9\u4e2a\u4e0d\u4e00\u6837\u7684\u8bdd\u5e0c\u671b\u5927\u5bb6\u5206\u4eab\u4e00\u4e0b\u3002
\n#include<iostream>
\n#include<cstdio>
\n#include<cstring>
\nusing namespace std;<\/p>\n
int a[1000005];<\/p>\n
\/*struct F
\n{
\nint a,i;
\n}*\/
\nint f[3000005];<\/p>\n
int main()
\n{
\nint n,k;
\nwhile(scanf(“%d%d”,&n,&k)>0)
\n{
\nfor(int i=0;i<n;i++)
\nscanf(“%d”,&a[i]);
\nint l=0,r=-1;
\nmemset(f,0,sizeof(f));
\nfor(int i=0;i<k;i++)
\n{
\nwhile(r>=l&&a[i]<=a[f[r]])r–;
\nf[++r]=i;
\n}
\nprintf(“%d”,a[f[0]]);
\nfor(int i=k;i<n;i++)
\n{
\nwhile(l<=r&&f[l]+k-1<i)l++;
\nwhile(r>=l&&a[i]<=a[f[r]])r–;
\nf[++r]=i;
\nprintf(” %d”,a[f[l]]);
\n}
\nprintf(“n”);
\nl=0,r=-1;
\nmemset(f,0,sizeof(f));
\nfor(int i=0;i<k;i++)
\n{
\nwhile(r>=l&&a[i]>=a[f[r]])r–;
\nf[++r]=i;
\n}
\nprintf(“%d”,a[f[0]]);
\nfor(int i=k;i<n;i++)
\n{
\nwhile(l<=r&&f[l]+k-1<i)l++;
\nwhile(r>=l&&a[i]>=a[f[r]])r–;
\nf[++r]=i;
\nprintf(” %d”,a[f[l]]);
\n}
\nprintf(“n”);
\n}
\nreturn 0;
\n}
\n\/**************************************************************
\nProblem: 1117
\nUser: admin
\nLanguage: C++
\nResult: Accepted
\nTime:307 ms
\nMemory:16884 kb
\n****************************************************************\/
\n\u590d\u5236\u4ee3\u7801<\/p>\n","protected":false},"excerpt":{"rendered":"
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