Matrix
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1309 Accepted Submission(s): 572
Problem Description
Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the '1' in this row or this column .
Your task is to give out the minimum times of deleting all the '1' in the matrix.
Input
There are several test cases.
The first line contains two integers n,m(1<=n,m<=100), n is the number of rows of the given matrix and m is the number of columns of the given matrix.
The next n lines describe the matrix:each line contains m integer, which may be either ‘1’ or ‘0’.
n=0 indicate the end of input.
Output
For each of the test cases, in the order given in the input, print one line containing the minimum times of deleting all the '1' in the matrix.
Sample Input
3 3
0 0 0
1 0 1
0 1 0
0
Sample Output
2
把行看做一个集合，把列看做一个集合
最小顶点覆盖=最大匹配
1 #include<cstring> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstdio> 6 using namespace std; 7 8 int link[1001][1001]; 9 int cx[1001]; 10 int cy[1001]; 11 int mk[1001]; 12 13 void init() 14 { 15 memset(link,0,sizeof(link)); 16 memset(cx,0xff,sizeof(cx)); 17 memset(cy,0xff,sizeof(cy)); 18 memset(mk,0,sizeof(mk)); 19 } 20 int nx,ny; 21 22 int path(int u) 23 { 24 int v; 25 for(v=1;v<=ny;v++) 26 { 27 if(!mk[v]&&link[u][v]) 28 { 29 mk[v]=1; 30 if(cy[v]==-1||path(cy[v])) 31 { 32 cx[u]=v; 33 cy[v]=u; 34 return 1; 35 } 36 } 37 } 38 return 0; 39 } 40 41 int maxmatch() 42 { 43 int i; 44 int sum=0; 45 for(i=1;i<=nx;i++) 46 { 47 if(cx[i]==-1) 48 { 49 memset(mk,0,sizeof(mk)); 50 sum+=path(i); 51 } 52 } 53 return sum; 54 } 55 56 int main() 57 { 58 while(scanf("%d",&nx)!=EOF) 59 { 60 if(!nx)break; 61 init(); 62 scanf("%d",&ny); 63 int i,j; 64 for(i=1;i<=nx;i++) 65 { 66 for(j=1;j<=ny;j++) 67 { 68 scanf("%d",&link[i][j]); 69 } 70 } 71 printf("%d ",maxmatch()); 72 } 73 return 0; 74 }