HDU-2732 Leapin' Lizards

题目链接：HDU-2732 Leapin' Lizards

题意

给出$n imes m$的网格，网格上的一些位置上有一只蜥蜴，所有蜥蜴的最大跳跃距离是$d$，如果一只蜥蜴能跳出网格，那么它就安全了，且每个格子有一个最大跳出次数$x$，即如果有$x$只蜥蜴从这个格子跳出，这个格子就再也不能有蜥蜴进来了，问最少有多少只蜥蜴跳不出网格。

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思路

设源点为$s$，汇点为$t$，建图如下：

$x=0$的格子为无用格子，全部不用考虑；

对于一个格子$u$，将其拆成两个结点，入点$u$和出点$u'$；

若一个格子$u$上有蜥蜴，那么连边$(s, u, 1)$；

若格子$u$能承受$x$次跳出，那么连边$(u, u', x)$；

若格子$u$能一次跳出网格，那么连边$(u', t, inf)$；

对于每对格子$u,v(u eq v)$，$u,v$的曼哈顿距离$leq d$，那么连边$(u', v, inf),(v', u, inf)$。

最大流就是能跳出网格的蜥蜴的最大数量。

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代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
using std::queue;
const int INF = 0x3f3f3f3f, N = 2000, M = 200000;
int head[N], d[N];
int s, t, tot, maxflow;
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
    edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}
void init(int v_num) {
    tot = 1, maxflow = 0;
    s = v_num, t = s + 1;
    memset(head, 0, sizeof(head));
}

int main() {
    int T, cas = 0;
    scanf("%d", &T);
    while (T--) {
        int n, m, dd;
        char str[25][25];
        scanf("%d %d", &n, &dd);
        for (int i = 0; i < n; i++) {
            scanf(" %s", str[i]);
            if (i == 0) {
                m = strlen(str[0]);
                init(n * m * 2);
            }
            for (int j = 0, u, v; j < m; j++) if (str[i][j] - '0' > 0) {
                u = i * m + j;
                add(u, u + n * m, str[i][j] - '0');
                if (i < dd || i + dd >= n || j < dd || j + dd >= m) {
                    add(u + n * m, t, INF);
                }
                for (int k = 0; k <= i; k++) for (int p = 0; p < m; p++) {
                    if (k == i && p >= j) break;
                    if (str[k][p] - '0' > 0) {
                        v = k * m + p;
                        if (abs(i - k) + abs(j - p) <= dd) {
                            add(u + n * m, v, INF);
                            add(v + n * m, u, INF);
                        }
                    }
                }
            }
        }
        int sum = 0;
        for (int i = 0; i < n; i++) {
            scanf(" %s", str[i]);
            for (int j = 0; j < m; j++) {
                if (str[i][j] == 'L') {
                    ++sum;
                    add(s, i * m + j, 1);
                }
            }
        }
        while (bfs()) maxflow += dinic(s, INF);
        sum -= maxflow;
        if (sum == 0) printf("Case #%d: no lizard was left behind.
", ++cas);
        else if (sum == 1) printf("Case #%d: 1 lizard was left behind.
", ++cas);
        else printf("Case #%d: %d lizards were left behind.
", ++cas, sum);
    }
    return 0;
}