标签：column div pac output cin 交换 find ons push
InputThere 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.OutputFor 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：给一个只有0 1的矩阵，是否能通过交换两行或者两列使对角线全为1
题解：二分图匹配x和y，输出方法是关键，两遍循环取对应的xy不相同的进行交换记录交换的行或者列（由矩阵知识可知行和列交换一种就行了）
刚开始因为but M should be more than 1000. 这句话我非要作死输出1000个，话说题目能不能不要瞎写啊！！！
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define pi acos(-1) #define ll long long #define mod 1000000007 using namespace std; const int N=100+5,maxn=100000+5,inf=0x3f3f3f3f; int color[N],n; bool used[N],ok[N][N]; bool match(int x) { for(int i=1;i<=n;i++) { if(ok[x][i]&&!used[i]) { used[i]=1; if(color[i]==0||match(color[i])) { color[i]=x; return 1; } } } return 0; } int solve() { int ans=0; memset(color,0,sizeof color); for(int i=1;i<=n;i++) { memset(used,0,sizeof used); ans+=match(i); } return ans; } int main() { ios::sync_with_stdio(false); cin.tie(0); while(cin>>n){ for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cin>>ok[i][j]; if(solve()!=n)cout<<-1<<endl; else { // for(int i=1;i<=n;i++)cout<<color[i]<<endl; memset(used,0,sizeof used); queue<pair<int,int> >q; for(int i=1;i<=n;i++) { if(i==color[i])continue; for(int j=i+1;j<=n;j++) { if(j==color[j])continue; if(i==color[j]) { q.push(make_pair(i,j)); swap(color[i],color[j]); } } } cout<<q.size()<<endl; while(!q.empty()){ cout<<"C "<<q.front().first<<" "<<q.front().second<<endl; q.pop(); } } } return 0; }
标签：column div pac output cin 交换 find ons push
原文地址：http://www.cnblogs.com/acjiumeng/p/6740168.html
