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;
}