FZU 1202 信与信封问题 二分图匹配

找匹配中的关键边。

做法： 拆掉一条匹配边，然后对边两边的点做一次增广，如果可以增广，那么此边不是关键边，否则是关键边。

 详情可以参见：http://www.docin.com/p-109868135.html

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <time.h>

using namespace std;

const int INF = 1<<30;
const int MAXN = 105;


bool Matrix[MAXN][MAXN];
int n;

int mX[MAXN], mY[MAXN], visX[MAXN], visY[MAXN];

bool dfs(int u) {
    visX[u] = true;
    for (int v = 1; v <= n; v++) if (!Matrix[u][v] && !visY[v]) {
        visY[v] = true;
        if (mY[v]<0 || dfs(mY[v])) {
            mY[v] = u;
            mX[u] = v;
            return true;
        }
    }
    return false;
}

void input() {
    memset(Matrix, false, sizeof(Matrix));
    int x, y;
    while (scanf("%d%d", &x, &y)!=EOF && (x!=0||y!=0)) {
        Matrix[x][y] = true;
    }
}

void solve() {
    int Max;
    memset(mX, -1, sizeof(mX));
    memset(mY, -1, sizeof(mY));
    for (int i = 1; i <= n; i++) {
        memset(visX, false, sizeof(visX));
        memset(visY, false, sizeof(visY));
        if (dfs(i)) Max++;
    }
    if (Max<n) {
        puts("none");
        return ;
    }
    bool flag = true;
    int delX, delY;
    for (int i = 1; i <= n; i++) {
        delX = i; delY = mX[i];
        mX[delX] = mY[delY] = -1;
        Matrix[delX][delY] = true;
        memset(visX, false, sizeof(visX));
        memset(visY, false, sizeof(visY));

        if (!dfs(i)) {
            printf("%d %d
", delX, delY);
            mX[delX] = delY; mY[delY] = delX;
            flag = false;
        }
        Matrix[delX][delY] = false;
    }
    if (flag) puts("none");
}

int main() {
    #ifdef Phantom01
        freopen("FZU1202.txt", "r", stdin);
    #endif //Phanaom01

    while (scanf("%d", &n)!=EOF) {
        input();
        solve();
        puts("");
    }

    return 0;
}