洛谷 P3191 [HNOI2007]紧急疏散EVACUATE（网络最大流）

题解

二分答案+Dinic最大流

二分答案\(mid\)

把门拆成\(mid\)个时间点的门

相邻时间的门连一条\(inf\)的边

预处理出每个门到每个人的最短时间

为\(dis[k][i][j]\) 在\((i,j)\)的人到第\(k\)个门最短时间

然后一个人连向每个第\(dis[k][i][j]\)那个时刻的门，容量为\(1\)

然后,源点连向每个人一条容量为\(1\)的边

所有门都连向汇点一条容量为\(1\)的边（其实只要每个最后一个时刻的门连一条容量为\(mid\)的边即可）

Code

#include<bits/stdc++.h>
#define LL long long
#define RG register
using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 40, inf = 1e9;
int n, m;
int fx[4] = {1, 0, -1, 0};
int fy[4] = {0, 1, 0, -1};
char ch[N][N];
int dis[N<<2][N][N], door, people;
struct BB {
    int x, y;
};
queue<BB> Q;
void BFS(int k, int xx, int yy) {
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            dis[k][i][j] = inf;
    dis[k][xx][yy] = 0; Q.push((BB) {xx, yy});
    while (!Q.empty()) {
        int x = Q.front().x, y = Q.front().y; Q.pop();
        for (int w = 0; w < 4; w++) {
            int i = x + fx[w], j = y + fy[w];
            if (i < 1 || i > n || j < 1 || j > m || ch[i][j] != '.') continue;
            if (dis[k][i][j] > dis[k][x][y]+1)
                dis[k][i][j] = dis[k][x][y]+1, Q.push((BB) {i, j});
        }
    }
}
struct node {
    int to, nxt, w;
}g[N*N*N*N];
int last[N*N*N], cur[N*N*N], dep[N*N*N], s, t, gl = 1, p[N][N]; 
inline void add(int a, int b, int c) {
    g[++gl] = (node) {b, last[a], c};
    last[a] = gl;
    g[++gl] = (node) {a, last[b], 0};
    last[b] = gl;
}
queue<int> q;
bool bfs() {
    memset(dep, 0, sizeof(dep));
    q.push(s); dep[s] = 1;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = last[u]; i; i = g[i].nxt) {
            int v = g[i].to;
            if (!dep[v] && g[i].w) {
                dep[v] = dep[u]+1;
                q.push(v);
            }
        }
    }
    return dep[t] == 0 ? 0 : 1;
}
int dfs(int u, int d) {
    if (u == t) return d;
    for (int &i = cur[u]; i; i = g[i].nxt) {
        int v = g[i].to;
        if (g[i].w && dep[v] == dep[u]+1) {
            int di = dfs(v, min(d, g[i].w));
            if (di) {
                g[i].w -= di;
                g[i^1].w += di;
                return di;
            }
        }
    }
    return 0;
}
int Dinic() {
    int ans = 0;
    while (bfs()) {
        for (int i = 1; i <= t; i++)
            cur[i] = last[i];
        while (int d = dfs(s, inf)) ans += d;
    }
    return ans;
}		 
int check(int mid) {
    s = door*mid+people+1, t = s+1;
    memset(last, 0, sizeof(last)); gl = 1;
    for (int k = 1; k <= door; k++)
        for (int z = 1; z <= mid; z++) {
            if (z < mid)
                add((k-1)*mid+z, (k-1)*mid+z+1, inf);
            else add(k*mid, t, mid);
        }
    for (int i = 1; i <= people; i++)
        add(s, door*mid+i, 1);
    for (int k = 1; k <= door; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                if (ch[i][j] == '.' && dis[k][i][j] <= mid)
                    add(door*mid+p[i][j], (k-1)*mid+dis[k][i][j], 1);
    return Dinic();
}
int main() {
    read(n), read(m);
    for (int i = 1; i <= n; i++)
        scanf("%s", ch[i]+1);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (ch[i][j] == 'D')
                BFS(++door, i, j);
            else if (ch[i][j] == '.') p[i][j] = ++people;
    int l = 0, r = people, ans = 666666;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid) == people) r = mid-1, ans = mid;
        else l = mid+1;
    }
    if (ans == 666666) puts("impossible");
    else printf("%d\n", ans);
    return 0;
}