HDU3360 - National Treasures

二分匹配求最小点覆盖

因为宝藏防止保安的位置在与其奇偶性不同的位置，所以分奇偶建图，如果同时存在宝藏就必定有一点为保安

#include <cstring>
#include <iostream>
#include <cstdlib>
#include <cctype>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#define INF (int)(1e9)
#define maxn 5010
using namespace std;
typedef long long ll;
int p = 0;
struct node {
    int to, w, next;
};
node edge[maxn * maxn];
int head[maxn];
void add_edge(int u, int v, int w) {
    edge[p].to = v;
    edge[p].w = w;
    edge[p].next = head[u];
    head[u] = p;
    ++ p;
}
int Mx[maxn],My[maxn],uN,vN;
int dx[maxn],dy[maxn],dis;
bool vis[maxn];
bool search_path() {
    queue<int> Q;
    dis = INF;
    memset(dx, -1, sizeof(dx));
    memset(dy, -1, sizeof(dy));
    for (int i = 0; i < uN; i++)
        if(Mx[i] == -1) {
            Q.push(i);
            dx[i] = 0;
        }
    while (!Q.empty()) {
        int u = Q.front();
        Q.pop();
        if (dx[u] > dis)  break;
        int v;
        for (int k = head[u]; k != -1; k = edge[k].next) {
            v = edge[k].to;
            if (dy[v] == -1) {
                dy[v] = dx[u] + 1;
                if (My[v] == -1)  dis = dy[v];
                else {
                    dx[My[v]] = dy[v]+1;
                    Q.push(My[v]);
                }
            }
        }
    }
    return dis != INF;
}
bool dfs(int u) {
    int v;
    for (int k = head[u]; k != -1; k = edge[k].next) {
        v = edge[k].to;
        if (!vis[v] && dy[v]==dx[u]+1) {
            vis[v] = 1;
            if (My[v] != -1 && dy[v] == dis) continue;
            if(My[v] == -1 || dfs(My[v])) {
                My[v] = u;
                Mx[u] = v;
                return true;
            }
        }
    }
    return false;
}
int max_match() {
    int res = 0;
    memset(Mx, -1, sizeof(Mx));
    memset(My, -1, sizeof(My));
    while (search_path()) {
        memset(vis, 0, sizeof(vis));
        for (int i = 0; i < uN; i ++)
            if (Mx[i] == -1 && dfs(i))  res ++;
    }
    return res;
}
int mat[maxn][maxn];
int dir[12][2] = {-1, -2, -2, -1, -2, 1, -1, 2, 1, 2, 2, 1, 2, -1, 1, -2, -1, 0, 0, 1, 1, 0, 0, -1};
int even[maxn], odd[maxn];
int main() {
    int n, m, now = 0;
    while (scanf("%d%d", &n, &m) != EOF) {
        memset(head, -1, sizeof(head)), p = 0;
        if (n == 0 && m == 0) break;
        int p1 = 0, p2 = 0;
        for (int i = 0; i < n; ++ i) {
            for (int j = 0; j < m; ++ j) {
                scanf("%d", &mat[i][j]);
                int tmp = i*m+j;
                if ((i+j)%2) {
                    odd[tmp] = p1++;
                }
                else {
                    even[tmp] = p2++;
                }
            }
        }
        for (int i = 0; i < n; ++ i) {
            for (int j = 0; j < m; ++ j) if (mat[i][j] != -1) {
                int x = mat[i][j];
                for (int k = 0; k < 12; ++ k, x >>= 1) if (x&1) {
                    int nx = i + dir[k][0];
                    int ny = j + dir[k][1];
                    int tmp = i*m+j;
                    if (mat[nx][ny] != -1 && nx >= 0 && nx < n && ny >= 0 && ny < m) {
                        if ((i+j)%2) {
                            add_edge(odd[tmp],even[nx*m+ny],0);
                        }
                        else {
                            add_edge(odd[nx*m+ny],even[tmp],0);
                        }

                    }
                }
            }
        }
        uN = p1, vN = p2;
        printf("%d. ", ++ now);
        cout << max_match() << endl;
    }
}