HDOJ 题目2189 Swap（二分图最大匹配，输出路径）

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Swap

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1789 Accepted Submission(s): 598
Special Judge

Problem Description

Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?

Input

There are several test cases in the input. The first line of each test case is an integer N (1 <= N <= 100). Then N lines follow, each contains N numbers (0 or 1), separating by space, indicating the N*N matrix.

Output

For each test case, the first line contain the number of swaps M. Then M lines follow, whose format is “R a b” or “C a b”, indicating swapping the row a and row b, or swapping the column a and column b. (1 <= a, b <= N). Any correct answer will be accepted, but M should be more than 1000.

If it is impossible to make all the diagonal entries equal to 1, output only one one containing “-1”.

Sample Input

20 11 021 01 0

Sample Output

1R 1 2-1

Source

2009 Multi-University Training Contest 1 - Host by TJU

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ac代码

#include<stdio.h>#include<string.h>int map[1010][1010],r[1010][1010],link[1010],vis[1010],n,row1[1010],row2[1010];int dfs(int u){int i;for(i=1;i<=n;i++){if(map[u][i]&&!vis[i]){vis[i]=1;if(link[i]==-1||dfs(link[i])){link[i]=u;return 1;}}}return 0;}int main(){//int n;while(scanf("%d",&n)!=EOF){int i,j;for(i=1;i<=n;i++){for(j=1;j<=n;j++)scanf("%d",&map[i][j]);}memset(link,-1,sizeof(link));memset(r,0,sizeof(r));for(i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(!dfs(i))break;}if(i<=n)printf("-1\n");else{int k=0;for(i=1;i<=n;i++){if(r[i][link[i]]==0&&i!=link[i]){k++;r[i][link[i]]=r[link[i]][i]=1;if(i<link[i]){row1[k]=i;row2[k]=link[i];}else{row1[k]=link[i];row2[k]=i;}for(j=i+1;j<=n;j++){if(link[j]==i){link[j]=link[i];break;}}}}printf("%d\n",k);for(i=1;i<=k;i++){printf("R %d %d\n",row1[i],row2[i]);}}}}

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