题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2819
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
2 0 1 1 0 2 1 0 1 0
Sample Output
1 R 1 2 -1
二分图左边的节点为每一行的行号
二分图右边的节点为每一行中出现的“1”对应的列号如果某行或者某列全是0。那怎么换都没办法的。
否则,一定能换出来。其次要明确:只交换行或者只交换列都是可以换出来的。
#include<cstring> #include<iostream> #include<algorithm> #include<cmath> #include<queue> #define INF 99999999; using namespace std; int n; int mp[110][110]; int match[110]; int vis[110]; int dfs(int u) { for (int v = 1; v <= n; v++) { if (mp[u][v] && !vis[v]) { vis[v] = 1; if (match[v] == -1 || dfs(match[v])) { match[v] = u; return 1; } } } return 0; } int hungarian() { int ans = 0; memset(match, -1, sizeof(match)); for (int i = 1; i <= n; i++) { memset(vis, 0, sizeof(vis)); ans += dfs(i); } return ans; } int main() { //freopen("C://input.txt", "r", stdin); while (scanf("%d", &n)!=EOF) { memset(mp, 0, sizeof(mp)); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) scanf("%d", &mp[i][j]); if (hungarian() != n) printf("-1\n"); else { int num = 0; int a[110], b[110]; for (int i = 1; i <= n; i++) { if (i != match[i]) { for (int j = 1; j <= n; j++) { if (i == match[j]) { a[num] = i; b[num] = j; num++; swap(match[i], match[j]); break; } } } } printf("%d\n", num); for (int i = 0; i < num; i++) { printf("C %d %d\n", a[i], b[i]); } } } return 0; }
hdu 2819 Swap (二分图最大匹配)
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转载自blog.csdn.net/Evildoer_llc/article/details/83021208
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