FOJ 1467 Pick Up Points

Problem 1467 Pick Up Points

Accept: 163    Submit: 583
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

 

Recently loy are very interested in playing a funny game. There is a board of n*m grid. Each grid has one point or not. If two adjacency grids both have one point, someone can pick them up, and then these two grids have no point. The two grids are adjacency only if the Manhattan distence between them is exactly 1.

The purpose of the game is to gain as many points as possible.

Input

The first line is an interger T, means the number of test cases. Each test case begins with two interger n,m (1 <= n,m <= 100), means the size of the board is n*m. Next n line, each line has m intergers which is 1 or 0, means the grid has one point or not.

Output

Output the maximum point loy can get.

Sample Input

2

3 4

0 1 0 1

0 1 1 0

1 1 1 1

2 2

1 0

0 1

Sample Output

6

0

Source

LOY

// 用了匈牙利算法，原本以爲會超時,沒想到46ms就過了,哈哈，暫列第四位。 #include<stdio.h> int arc[101][101]; int link[101][101]; int visited[101][101]; int dir[4][2]={-1,0,1,0,0,-1,0,1}; int n,m,p; int ko(int a,int b) { if(a>=0 && b>=0 && a<n && b<m && visited[a][b]==0 && arc[a][b]==1) return 1; return 0; } struct node // 記錄訪問過的點 { int x,y; }cc[10000]; int dfs(int a,int b) { int tx,ty; int i; for(i=0;i<4;i++) { tx=a+dir[i][0]; ty=b+dir[i][1]; if(ko(tx,ty)) { visited[tx][ty]=1; cc[p].x=tx; cc[p++].y=ty; if(link[tx][ty]==-1) { link[tx][ty]=i; // 這裏賦值i,表示是從加a+dir[i][0],及b+dir[i][1] 得到的,即指向和它組和的點 return 1; } else if(dfs(tx-dir[link[tx][ty]][0],ty-dir[link[tx][ty]][1])) // 尋找與它組合的點的增廣路徑 { link[tx][ty]=i; return 1; } } } return 0; } int main() { int t,i,j,k; int a,b; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=0;i<n;i++) for(j=0;j<m;j++) { scanf("%d",&arc[i][j]); link[i][j]=-1; visited[i][j]=0; } p=0; for(i=k=0;i<n;i++) { for(j=0;j<m;j++,p=0) { if(arc[i][j]==1) { if(dfs(i,j)) k++; } for(a=0;a<p;a++) // 只將訪問過的點全部置回未標記 visited[cc[a].x][cc[a].y]=0; } } printf("%d/n",k); } return 0; }