Swap
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3102 Accepted Submission(s): 1117
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
2
0 1
1 0
2
1 0
1 0
Sample Output
1
R 1 2
-1
题意:给定一个n*n的01矩阵,要求通过若干次行交换或列交换来满足主对角线上的数字均为1。
首先能确定的一点,只用行交换或只用列交换就能满足。一开始看到跟覆盖有关,还在想会不会是舞蹈链什么的,后来发现并不需要。可以把对角线上某一点为1理解为行与列的匹配,这样行和列就正好构成了一个二分图。如果这个二分图的最大匹配为n的话就存在方案,此时,根据匈牙利算法的匹配数组就能找到解法。
#include <stdio.h> #include <string.h> int a[105], b[105]; class Hungary { #define Hungary_MAX_Node 105 #define Hungary_MAX_Edge 10005 public: struct EDGE { int v; int next; } edge[Hungary_MAX_Edge]; int head[Hungary_MAX_Node]; int Left[Hungary_MAX_Node]; bool vis[Hungary_MAX_Node]; int N, M; Hungary() { clear(); } void clear() { N = M = 0; memset(Left, 0, sizeof(Left)); memset(head, -1, sizeof(head)); } void addEdge(int a, int b) { edge[M].v = b; edge[M].next = head[a]; head[a] = M++; } bool dfs(int u) { for (int e = head[u]; e != -1; e = edge[e].next) { int v = edge[e].v; if (!vis[v]) { vis[v] = true; if (!Left[v] || dfs(Left[v])) { Left[v] = u; return true; } } } return false; } int Max_Match() { int ret = 0; for (int i = 1; i <= N; i++) { memset(vis, 0, sizeof(vis)); if (dfs(i)) { ret++; } } return ret; } }; Hungary hungary; int main() { int N; while (~scanf("%d", &N)) { hungary.clear(); hungary.N = N; int x, ans; for (int i = 1; i <= N; i++) for (int j = 1; j <= N; j++) { scanf("%d", &x); if (x) { hungary.addEdge(i, j); } } ans = hungary.Max_Match(); if (ans < N) { printf("-1 "); continue; } int tot = 0, j; for (int i = 1; i <= N; i++) { for (j = 1; j <= N && hungary.Left[j] != i; j++); if (i != j) { a[tot] = i; b[tot] = j; tot++; int t = hungary.Left[i]; hungary.Left[i] = hungary.Left[j]; hungary.Left[j] = t; } } printf("%d ", tot); for (int i = 0; i < tot; i++) { printf("C %d %d ", a[i], b[i]); } } return 0; }