Swap
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3800 Accepted Submission(s): 1401Special 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
Source
按行交换。对行进行1-n编号，对列也进行编号。
若第i行的第j1、j2……列有1，则节点i与j1、j2……连边。
二分图跑最大匹配，为完美匹配方可。
matching数组保存每个点的匹配，模拟一下交换输出答案。
1 //2017-08-26 2 #include3 #include 4 #include 5 #include 6 7 using namespace std; 8 9 const int N = 100000; 10 const int M = 5000000; 11 int head[N], tot; 12 struct Edge{ 13 int to, next; 14 }edge[M]; 15 16 void init(){ 17 tot = 0; 18 memset(head, -1, sizeof(head)); 19 } 20 21 void add_edge(int u, int v){ 22 edge[tot].to = v; 23 edge[tot].next = head[u]; 24 head[u] = tot++; 25 26 edge[tot].to = u; 27 edge[tot].next = head[v]; 28 head[v] = tot++; 29 } 30 31 int n; 32 int matching[N]; 33 int check[N]; 34 bool dfs(int u){ 35 for(int i = head[u]; i != -1; i = edge[i].next){ 36 int v = edge[i].to; 37 if(!check[v]){ //要求不在交替路 38 check[v] = 1;//放入交替路 39 if(matching[v] == -1 || dfs(matching[v])){ 40 //如果是未匹配点，说明交替路为增广路，则交换路径，并返回成功 41 matching[u] = v; 42 matching[v] = u; 43 return true; 44 } 45 } 46 } 47 return false;//不存在增广路 48 } 49 50 //hungarian: 二分图最大匹配匈牙利算法 51 //input: null 52 //output: ans 最大匹配数 53 int hungarian(){ 54 int ans = 0; 55 memset(matching, -1, sizeof(matching)); 56 for(int u = 1; u <= n; u++){ 57 if(matching[u] == -1){ 58 memset(check, 0, sizeof(check)); 59 if(dfs(u)) 60 ans++; 61 } 62 } 63 return ans; 64 } 65 66 const int MAXID = 100; 67 int a[N], b[N], cnt; 68 69 int main() 70 { 71 std::ios::sync_with_stdio(false); 72 //freopen("inputE.txt", "r", stdin); 73 while(cin>>n){ 74 init(); 75 int v; 76 for(int i = 1; i <= n; i++){ 77 for(int j = 1; j <= n; j++){ 78 cin>>v; 79 if(v){ 80 add_edge(i, j+MAXID); 81 } 82 } 83 } 84 int match = hungarian(); 85 if(match != n)cout<<-1<