SGU 190.Dominoes(二分图匹配)

时间限制：0.25s

空间限制：4M

题意:

      给定一个N*N的棋盘，一些格子被移除，在棋盘上放置一些1*2的骨牌，判定能否放满，并且输出任意方案。

 

 

 

 

 

 

 

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Solution：

             首先考虑对棋盘的一个格子黑白染色（实际上不需要），得到一个类似国际象棋棋盘的东西，一个骨牌能放置在相邻的一对黑白格子上

             我们考虑对每一个黑格子，连一条到相邻白色格子的边，然后做二分图的最大匹配，判断是否是完备匹配，输出解即可。

 　　　　思路比较简单直接，输出需要一些简单技巧和小处理。

 

code

#include <iostream>

#include <cstring>

#include <fstream>

#include <cmath>

#include <cstdio>

using namespace std;

const int INF = 1700;

struct node {

    int u, v, next;

} edge[100000];

int pHead[INF], vis[INF], pr[INF];

int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};

int n, m, x, y, nCnt, an;

int exPath (int x) {

    for (int k = pHead[x]; k != 0; k = edge[k].next) {

        int x = edge[k].u, y = edge[k].v;

        if (!vis[y]) {

            vis[y] = 1;

            if ( !pr[y] || exPath (pr[y]) ) return pr[y] = x;

        }

    }

    return 0;

}

void addEdge (int u, int v) {

    edge[++nCnt].u = u, edge[nCnt].v = v;

    edge[nCnt].next = pHead[u];

    pHead[u] = nCnt;

}

int g[50][50];

int main() {

       //ofstream cout("out.txt");

    cin >> n >> m;

    for (int i = 1; i <= n; i++)

        for (int j = 1; j <= n; j++) g[i][j] = 1;

    for (int i = 1; i <= m; i++) {

        cin >> x >> y;

        g[x][y] = 0;

    }

    for (int i = 1; i <= n; i++)

        for (int j = 1; j <= n; j++) {

            if (g[i][j])

                for (int k = 0; k < 4; k++) {

                    int x = i + dx[k], y = j + dy[k];

                    if (g[x][y])

                        addEdge ( (i - 1) *n + j, (x - 1) *n + y);

                }

        }

    for (int i = 1; i <= n * n; i++) {

        if (exPath (i) ) an++;

        memset (vis, 0, sizeof vis);

    }

    int t1 = 0, t2 = 0;

    int ans[2][INF];

    for (int i = 1; i <= n * n; i++) {

        if (pr[i] && !vis[i]) {

            vis[i] = vis[pr[i]] = 1;

            if (abs (pr[i] - i) == n)

                ans[0][++t1] = min (i, pr[i]);

            else

                ans[1][++t2] = min (i, pr[i]);

        }

    }

    if (an == (n * n - m) ) {

        cout << "Yes" << endl;

        cout << t1 << endl;

        for (int i = 1; i <= t1; i++) {

            int l, r;

            if (ans[0][i] % n) l = ans[0][i] / n + 1, r = ans[0][i] % n;

            else

                l = ans[0][i] / n, r = n;

            cout << l << ' ' << r << endl;

        }

        cout << t2 << endl;

        for (int i = 1; i <= t2; i++) {

            int l, r;

            l = ans[1][i] / n + 1, r = ans[1][i] % n;

            cout << l << ' ' << r << endl;

        }

    }

    else

        cout << "No";

    return 0;

}

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