洛谷 P3227 [HNOI2013]切糕（最小割）

题解

Dinic求最小割

题目其实就是求最小的代价使得每个纵轴被分成两部分

最小割！！！

我们把每个点抽象成一条边，一个纵轴就是一条\(S-T\)的路径

但是题目要求\(|f(x,y)-f(x’,y’)| ≤D\)
不能直接跑最小割

考虑如何限制

首先，\(|f(x,y)-f(x’,y’)| ≤D\)是相互的

所以只要考虑 \(f(x,y)-f(x',y'）\leq D\)

然后，我们需要强大的 图解

代码就很简洁了

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 = 80000, inf = 2147483647;

struct node {
    int to, nxt, w;
}g[2000000];
int last[N], gl = 1;

void add(int x, int y, int z) {
    g[++gl] = (node) {y, last[x], z};
    last[x] = gl;
    g[++gl] = (node) {x, last[y], 0};
    last[y] = gl;
}

queue<int> q;
int dep[N], s, t, cur[N];

bool bfs() {
    memset(dep, 0, sizeof(dep));
    dep[s] = 1;
    q.push(s);
    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 (dep[v] == dep[u]+1 && g[i].w) {
            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 a[50][50][50], id[50][50][50];

int fx[] = {0, 1, -1, 0};
int fy[] = {1, 0, 0, -1};

int main() {
    int p, q, r, d, tot = 0;
    read(p), read(q), read(r), read(d);
    for (int i = 1; i <= r; i++)
        for (int j = 1; j <= p; j++)
            for (int k = 1; k <= q; k++)
                read(a[i][j][k]), id[i][j][k] = ++tot;
    for (int j = 1; j <= p; j++)
        for (int k = 1; k <= q; k++)
            id[r+1][j][k] = ++tot;          
    s = tot+1, t = s+1;
    for (int i = 1; i <= p; i++)
        for (int j = 1; j <= q; j++)
            add(s, id[1][i][j], inf), add(id[r+1][i][j], t, inf);
    
    for (int k = 1; k <= r; k++)
        for (int i = 1; i <= p; i++)
            for (int j = 1; j <= q; j++)
                add(id[k][i][j], id[k+1][i][j], a[k][i][j]);
    
    for (int k = d+1; k <= r+1; k++)
        for (int i = 1; i <= p; i++)
            for (int j = 1; j <= q; j++) {
                for (int z = 0; z < 4; z++) {
                    int x = i + fx[z], y = j + fy[z];
                    if (x < 1 || y < 1 || x > p || y > q) continue;
                    add(id[k][i][j], id[k-d][x][y], inf);
                }
            }
    
    printf("%d\n", Dinic());
    return 0;
}