ZOJ-3581 a simple test

N * M matrix, where k grids have patterns. Q queries. If each query has a pattern on both grids, and it can be achieved by a maximum of two in disguise (there cannot be any other lattice with a pattern on the road). The pattern of the two grids will get two points, otherwise-1 points.

In fact, think carefully about the continuous watching game model. In the competition, I felt that the search was too violent to try it. In fact, the brute force writing was only 80 ms.

You can directly simulate brute-force attacks to meet the conditions.

#include <iostream>#include <cstdio>#include <queue>#include <cstdlib>#include <cstring>#include <iomanip>#include <algorithm>using namespace std;const int maxn=333;int map[maxn][maxn];int n,m;int k;int q;int ans;int x1,y1;int x2,y2;int fx[maxn],sx[maxn];int fy[maxn],sy[maxn];bool solve(){memset(fx,0,sizeof(fx));memset(sx,0,sizeof(sx));memset(fy,0,sizeof(fy));memset(sy,0,sizeof(sy));fx[x1]=1;sx[x2]=1;for(int i=x1+1;i<m;i++){if(map[i][y1]==0){fx[i]=1;}else{break;}}for(int i=x1-1;i>=0;i--){if(map[i][y1]==0){fx[i]=1;}else{break;}}for(int i=x2+1;i<m;i++){if(map[i][y2]==0){sx[i]=1;}else{break;}}for(int i=x2-1;i>=0;i--){if(map[i][y2]==0){sx[i]=1;}else{break;}}for(int i=0;i<m;i++){if(sx[i]&&fx[i]){bool flag=1;for(int j=min(y1,y2)+1;j<max(y1,y2);j++){if(map[i][j]){flag=0;break;}}if(flag){return true;}}}fy[y1]=1;sy[y2]=1;for(int i=y1+1;i<n;i++){if(map[x1][i]==0){fy[i]=1;}else{break;}}for(int i=y1-1;i>=0;i--){if(map[x1][i]==0){fy[i]=1;}else{break;}}for(int i=y2+1;i<n;i++){if(map[x2][i]==0){sy[i]=1;}else{break;}}for(int i=y2-1;i>=0;i--){if(map[x2][i]==0){sy[i]=1;}else{break;}}for(int i=0;i<n;i++){if(fy[i]&&sy[i]){bool flag=1;for(int j=min(x1,x2)+1;j<max(x2,x1);j++){if(map[j][i]){flag=0;break;}}if(flag){return true;}}}return false;}int main(){    while(scanf("%d%d",&m,&n)!=EOF)    {        ans=60;        memset(map,0,sizeof(map));        scanf("%d",&k);        for(int i=0;i<k;i++)        {            scanf("%d%d",&x1,&y1);            map[x1][y1]=1;        }        scanf("%d",&k);        while(k--)        {            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            if(map[x1][y1]==0||map[x2][y2]==0){ans--;}            else if(solve())            {                ans+=2;                map[x1][y1]=0;                map[x2][y2]=0;            }            else{ans--;}            printf("%d\n",ans);        }    }    return 0;}

ZOJ-3581 a simple test