Swap
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 972 Accepted Submission(s): 312
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
最大匹配
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<queue> 6 #include<cstring> 7 using namespace std; 8 9 int Link[1001][1001]; 10 int mk[1010]; 11 int cy[1010]; 12 int num[1001]; 13 int clo[1001]; 14 int cx[1011]; 15 int n; 16 struct node 17 { 18 int x,y; 19 }; 20 21 void init() 22 { 23 memset(Link,0,sizeof(Link)); 24 memset(mk,0,sizeof(mk)); 25 memset(num,0,sizeof(num)); 26 memset(cy,0xff,sizeof(num)); 27 memset(cx,0xff,sizeof(cx)); 28 memset(clo,0,sizeof(clo)); 29 } 30 31 int path(int u) 32 { 33 int v; 34 for(v=1;v<=n;v++) 35 { 36 if(!mk[v]&&Link[u][v]) 37 { 38 mk[v]=1; 39 if(cy[v]==-1||path(cy[v])) 40 { 41 cx[u]=v; 42 cy[v]=u; 43 return 1; 44 } 45 } 46 } 47 return 0; 48 } 49 50 int maxmatch() 51 { 52 int i; 53 int sum=0; 54 for(i=1;i<=n;i++) 55 { 56 if(cx[i]==-1) 57 { 58 memset(mk,0,sizeof(mk)); 59 sum+=path(i); 60 } 61 } 62 return sum; 63 } 64 65 int main() 66 { 67 while(scanf("%d",&n)!=EOF) 68 { 69 int i,j; 70 init(); 71 for(i=1;i<=n;i++) 72 { 73 for(j=1;j<=n;j++) 74 { 75 scanf("%d",&Link[i][j]); 76 } 77 } 78 int ans=maxmatch(); 79 if(ans<n){ 80 printf("-1 "); 81 continue; 82 } 83 queue<node>Q; 84 int sum=0; 85 for(i=1;i<=n;i++) 86 { 87 if(cx[i]!=i) 88 { 89 for(j=i+1;j<=n;j++) 90 { 91 if(cx[j]==i) 92 { 93 cx[j]=cx[i]; 94 cx[i]=i; 95 node t; 96 t.x=i,t.y=j; 97 Q.push(t); 98 sum++; 99 break; 100 } 101 } 102 } 103 } 104 printf("%d ",sum); 105 while(!Q.empty()) 106 { 107 node head=Q.front(); 108 Q.pop(); 109 printf("R %d %d ",head.x,head.y); 110 } 111 } 112 return 0; 113 }