【BZOJ3144】[Hnoi2013]切糕【最小割】

建图与最大流算法

最新推荐文章于 2022-01-08 11:18:56 发布

BraketBN

最新推荐文章于 2022-01-08 11:18:56 发布

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分类专栏：网络流

本文链接：https://blog.youkuaiyun.com/BraketBN/article/details/51490737

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本文详细介绍了一种基于最大流算法解决特定问题的方法。通过构建复杂的图结构，并利用深度优先搜索和广度优先搜索实现最大流求解，最终达到优化资源分配的目的。

【题目链接】

学习一发建图。

这篇题解比较详细【zarxdy34的题解】

/* Telekinetic Forest Guard */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxg = 45, maxn = 70005, maxm = 150005, maxq = 100000, inf = 0x3f3f3f3f;

int n, m, h, d, head[maxn], cur[maxn], cnt, bg, ed, depth[maxn], q[maxq], id[maxg][maxg][maxg];
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

struct _edge {
	int v, w, next;
} g[maxm << 1];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v, int w) {
	g[cnt] = (_edge){v, w, head[u]};
	head[u] = cnt++;
}

inline void insert(int u, int v, int w) {
	add(u, v, w); add(v, u, 0);
}

inline bool bfs() {
	for(int i = 0; i <= ed; i++) depth[i] = -1;
	int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1;
	while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) {
		depth[g[i].v] = depth[u] + 1;
		if(g[i].v == ed) return 1;
		q[t++] = g[i].v;
	}
	return 0;
}

inline int dfs(int x, int flow) {
	if(x == ed) return flow;
	int left = flow;
	for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) {
		int tmp = dfs(g[i].v, min(left, g[i].w));
		left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp;
		if(g[i].w) cur[x] = i;
		if(!left) return flow;
	}
	if(left == flow) depth[x] = -1;
	return flow - left;
}

int main() {
	n = iread(); m = iread(); h = iread(); d = iread();
	for(int i = 0; i < maxn; i++) head[i] = -1; cnt = 0;

	int tot = 0; bg = ++tot;
	for(int i = 1; i <= h + 1; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= m; k++)
		id[i][j][k] = ++tot;
	ed = ++tot;

	for(int i = 1; i <= h; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= m; k++) {
		int	w = iread();
		insert(id[i][j][k], id[i + 1][j][k], w);
		if(i + d <= h + 1) for(int r = 0; r < 4; r++) {
			int x = j + dx[r], y = k + dy[r];
			if(x < 1 || x > n || y < 1 || y > m) continue;
			insert(id[i + d][x][y], id[i][j][k], inf);
		}
	}
	for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) insert(bg, id[1][i][j], inf), insert(id[h + 1][i][j], ed, inf);

	int ans = 0;
	while(bfs()) {
		for(int i = 0; i <= ed; i++) cur[i] = head[i];
		ans += dfs(bg, inf);
	}

	printf("%d\n", ans);
	return 0;
}