Luogu P3355 骑士共存问题

最新推荐文章于 2022-11-13 00:07:59 发布

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最新推荐文章于 2022-11-13 00:07:59 发布

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原文链接：http://www.cnblogs.com/maomao9173/p/10479419.html

本文介绍了一种解决二分图最大独立集问题的算法，通过构建最大匹配来求解独立集。独立集问题是在图中寻找尽可能多的不相邻顶点集合，此算法利用了最大匹配与独立集之间的关系。

题目链接 \(Click\) \(Here\)

二分图最大独立集。对任意两个可以相互攻击的点，我们可以选其中一个。对于不会互相攻击的，可以全部选中。所以我们只需要求出最大匹配，根据定理，二分图最大独立集等于点数减去最大匹配，就得到了答案。

#include <bits/stdc++.h>
using namespace std;

const int N = 80010;
const int M = 800010;
const int INF = 0x3f3f3f3f;

int n, m, ban[210][210];
int cnt = -1, head[N];
int mv[8][2] = {{2, 1},{2, -1},{1, 2},{1, -2},{-2, 1},{-2, -1},{-1, 2},{-1, -2}};

bool in_map (int x, int y) {return 1 <= x && x <= n && 1 <= y && y <= n;}

struct edge {
    int nxt, to, f;
}e[M];

void add_edge (int from, int to, int flw) {
    e[++cnt].nxt = head[from];
    e[cnt].to = to;
    e[cnt].f = flw;
    head[from] = cnt;
}

void add_len (int u, int v, int f) {
    add_edge (u, v, f);
    add_edge (v, u, 0);
}

int nd1 (int x, int y) {return n * n * 0 + (x - 1) * n + y;}
int nd2 (int x, int y) {return n * n * 1 + (x - 1) * n + y;}

queue <int> q;
int cur[N], deep[N];

bool bfs (int s, int t) {
    memcpy (cur, head, sizeof (head));
    memset (deep, 0x3f, sizeof (deep));
    q.push (s); deep[s] = 0;
    while (!q.empty ()) {
        int u = q.front (); q.pop ();
        for (int i = head[u]; ~i; i = e[i].nxt) {
            int v = e[i].to;
            if (deep[v] == INF && e[i].f) {
                deep[v] = deep[u] + 1;
                q.push (v);
            }
        }
    }
    return deep[t] != INF;
}

int dfs (int u, int t, int lim) {
    if (u == t || !lim) {
        return lim;
    }
    int tmp = 0, flow = 0;
    for (int &i = cur[u]; ~i; i = e[i].nxt) {
        int v = e[i].to;
        if (deep[v] == deep[u] + 1) {
            tmp = dfs (v, t, min (lim, e[i].f));
            lim -= tmp;
            flow += tmp;
            e[i ^ 0].f -= tmp;
            e[i ^ 1].f += tmp;
            if (!lim) break;
        }
    }
    return flow;
}

int main () {
    memset (head, -1, sizeof (head));
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        int x, y;
        cin >> x >> y;
        ban[x][y] = true;
    }
    int s = n * n * 2 + 1, t = n * n * 2 + 2;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (ban[i][j]) continue;
            add_len (s, nd1 (i, j), 1);
            add_len (nd2 (i, j), t, 1);
            for (int k = 0; k < 8; ++k) {
                int tx = i + mv[k][0];
                int ty = j + mv[k][1];
                if (in_map (tx, ty) && !ban[tx][ty]) {
                    add_len (nd1 (i, j), nd2 (tx, ty), 1);
                }
            }
        }
    }
    int max_flow = 0;
    while (bfs (s, t)) {
        getchar ();
        max_flow += dfs (s, t, INF);
    }
    // printf ("max_flow = %d\n", max_flow);
    cout << n * n - m - max_flow / 2 << endl;
}

转载于:https://www.cnblogs.com/maomao9173/p/10479419.html