hdu 2819 Swap【完美二分匹配】

来源：互联网发布：java断点调试快捷键编辑：程序博客网时间：2024/05/18 03:22

Swap

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2514 Accepted Submission(s): 900
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

0 1

1 0

Sample Output

R 1 2

-1

Source

2009 Multi-University Training Contest 1 - Host by TJU

题目大意：给你一个 n*n的矩阵，使用行交换或者列交换的方式来使得矩阵的主对角线（左上到右下）都是1.

分析：对于这个题，不需要最少的操作次数，那么我们无论是怎样做，只要是搞定了这个问题，我们就胜利了，对于输出-1的情况，我们来行列匹配看看能否使得所有行都有列来匹配就好，如果没有，那么就不是完美匹配，也就是说输出-1，否则就一定有结果输出。

思路：

完美匹配：如果一个图的某个匹配中，所有的顶点都是匹配点，那么它就是一个完美匹配。图 4 是一个完美匹配。显然，完美匹配一定是最大匹配（完美匹配的任何一个点都已经匹配，添加一条新的匹配边一定会与已有的匹配边冲突）。但并非每个图都存在完美匹配。

遍历所有行，如果都有列匹配的话，就是一个完美匹配，否则输出-1.

如果是完美匹配，那么交换pri【】数组，使得pri【i】=i、

AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn = 505;int map[maxn][maxn];              //图的数组.int vis[maxn];                  //标记数组.int pri[maxn];int ans[maxn*2][2];int n;int  find(int x){    for(int i=1; i<=n; i++)    {        if(vis[i]==0&&map[i][x])        {            vis[i]=1;            if(pri[i]==-1||find(pri[i]))            {                pri[i]=x;                return 1;            }        }    }    return 0;}int main(){    while(~scanf("%d",&n))    {        memset(pri,-1,sizeof(pri));        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                scanf("%d",&map[i][j]);            }        }        int ok=1;        for(int i=1; i<=n; i++)        {            memset(vis,0,sizeof(vis));            if(find(i))continue;            else ok=0;        }        if(ok==0){printf("-1\n");continue;}        int cont=0;        for(int i=1;i<=n;i++)        {            if(pri[i]==i)continue;            for(int j=1;j<=n;j++)            {                if(pri[j]==i)                {                    swap(pri[i],pri[j]);                    ans[cont][0]=i;ans[cont++][1]=j;                }            }        }        printf("%d\n",cont);        for(int i=0;i<cont;i++)        {            printf("R %d %d\n",ans[i][0],ans[i][1]);        }    }}

0 0