[SCOI 2007] 蜥蜴

发表于2017-04-16更新于2020-08-06

题目大意

n×mn \times m 的网格上有一些柱子，一些柱子上有蜥蜴，蜥蜴可以跳向曼哈顿距离小于等于 dd 的柱子上，一个柱子上最多有一只蜥蜴，当蜥蜴跳走后，原柱子高度减一。求无法跳出网格的蜥蜴数量的最小值。

1⩽n,  m⩽201 \leqslant n, \; m \leqslant 20

1⩽d⩽41 \leqslant d \leqslant 4

题目链接

【SCOI 2007】蜥蜴 - Luogu 2472

题解

建图跑网络流，最大流为最多跳出去的蜥蜴只数。

每个点拆成两个点，记为 11 号点和 22 号点。

对于柱子高度的限制，从对应的 11 号点向 22 号点连一条容量为高度的边。

对与每一个点，从它的 22 号点向与其曼哈顿距离小于等于 dd 的点的 11 号点连一条容量无限的边。

最后，从源点向每一只蜥蜴的 11 号点连一条容量为 11 的边，从能跳出去的 22 号点向汇点连一条容量为11的边。

代码

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#include <cstdio>
#include <cstdlib>
#include <climits>
#include <queue>
#include <algorithm>
// #define DBG
const int MAXN = 20;
struct Edge;
struct Node {
    Edge *e, *curr;
    int level;
} N[MAXN * MAXN * 2 + 2];
struct Edge {
    Node *u, *v;
    Edge *next, *rev;
    int cap, flow;
    Edge(Node *u, Node *v, int cap) : u(u), v(v), cap(cap), flow(0), next(u->e) {}
};
void addEdge(int u, int v, int cap) {
#ifdef DBG
    printf("edge : %d --> %d, cap = %d\n", u, v, cap);
#endif
    N[u].e = new Edge(&N[u], &N[v], cap);
    N[v].e = new Edge(&N[v], &N[u], 0);
    N[u].e->rev = N[v].e;
    N[v].e->rev = N[u].e;
}
struct Dinic {
    bool makeLevelGraph(Node *s, Node *t, int n) {
        for (int i = 0; i < n; i++) N[i].level = 0;
        std::queue<Node *> q;
        q.push(s);
        s->level = 1;
        while (!q.empty()) {
            Node *u = q.front();
            q.pop();
            for (Edge *e = u->e; e; e = e->next) {
                if (e->cap > e->flow && e->v->level == 0) {
                    e->v->level = u->level + 1;
                    if (e->v == t) return true;
                    q.push(e->v);
                }
            }
        }
        return false;
    }
    int findPath(Node *s, Node *t, int limit = INT_MAX) {
        if (s == t) return limit;
        for (Edge *&e = s->curr; e; e = e->next) {
            if (e->cap > e->flow && e->v->level == s->level + 1) {
                int flow = findPath(e->v, t, std::min(limit, e->cap - e->flow));
                if (flow > 0) {
                    e->flow += flow;
                    e->rev->flow -= flow;
                    return flow;
                }
            }
        }
        return 0;
    }
    int operator()(int s, int t, int n) {
        int res = 0;
        while (makeLevelGraph(&N[s], &N[t], n)) {
            for (int i = 0; i < n; i++) N[i].curr = N[i].e;
            int flow;
            while ((flow = findPath(&N[s], &N[t])) > 0) res += flow;
        }
        return res;
    }
} dinic;
int n, m;
int getID(int x, int y, int k) {
    return k * n * m + (x - 1) * m + y;
}
int main() {
    int d;
    scanf("%d %d %d", &n, &m, &d);
    const int s = 0, t = n * m * 2 + 1;
    for (int i = 1; i <= n; i++) {
        static char str[MAXN + 1];
        scanf("%s", str + 1);
        for (int j = 1; j <= m; j++) if (str[j] != '0')
            addEdge(getID(i, j, 0), getID(i, j, 1), str[j] - '0');
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        static char str[MAXN + 1];
        scanf("%s", str + 1);
        for (int j = 1; j <= m; j++) if (str[j] == 'L') {
            cnt++;
            addEdge(s, getID(i, j, 0), 1);
        }
    }
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
        for (int k = 1; k <= n; k++) for (int l = 1; l <= m; l++) {
            if (abs(i - k) + abs(j - l) <= d) 
                addEdge(getID(i, j, 1), getID(k, l, 0), INT_MAX);
        }
        if (i <= d || i > n - d || j <= d || j > m - d) 
            addEdge(getID(i, j, 1), t, INT_MAX);
    }
    printf("%d\n", cnt - dinic(s, t, t + 1));
    return 0;
}