【最小费用最大流】JZOJ_4802 探险计划

题意

给出一个梯形，每个点上有一个危险值，从底部出发，每次能走到左上或者右上的点。

任务一：找出$m$条完全不相交的至底至顶的路径
任务二：找出$m$条仅在数字处相交的路径(可以重复经过点)

对于两个任务，求出最小的危险值总和。

思路

最小费用最大流。

将每个点拆点，对于任务一，给一个限制，任务二则无限制。

代码

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>

int n, m, tot, S, E, SS, sum, ans;
int a[81][161], ver[58001], next[58001], head[20004], flow[58001], cost[58001], d[20004], v[20004], pre[20004], p[81][161];

inline void add(int u, int v, int f, int c) {
	ver[++tot] = v;
	next[tot] = head[u];
	flow[tot] = f;
	cost[tot] = c;
	head[u] = tot;
	ver[++tot] = u;
	next[tot] = head[v];
	flow[tot] = 0;
	cost[tot] = -c;
	head[v] = tot;
}

inline int spfa() {
	std::queue<int> q;
	memset(d, 127 / 3, sizeof(d));
	memset(v, 0, sizeof(v));
	memset(pre, 0, sizeof(pre));
	d[SS] = 0;
	v[SS] = 1;
	q.push(SS);
	while (q.size()) {
		int u = q.front();
		q.pop();
		v[u] = 0;
		for (register int i = head[u]; i; i = next[i]) {
			if (!flow[i]) continue;
			if (d[ver[i]] > d[u] + cost[i]) {
				d[ver[i]] = d[u] + cost[i];
				pre[ver[i]] = i;
				if (!v[ver[i]]) {
					q.push(ver[i]);
					v[ver[i]] = 1;
				}
			}
		}
	}
	return d[E] < 707406378;
}

inline void addflow() {
    int i = E, mn = 2147483647;
    while (pre[i]) {
        mn = std::min(mn, flow[pre[i]]);
        i = ver[pre[i] ^ 1];
    }
    ans += mn * d[E];
	i = E;
    while (pre[i]) {
        flow[pre[i]] -= mn;
        flow[pre[i] ^ 1] += mn;
        i = ver[pre[i] ^ 1];
    }
}

int main() {
	scanf("%d %d", &n, &m);
	int tj = 0, sum = (m + m + n - 1) * n / 2;
	tot = 1;
	for (register int i = 1; i <= n; i++)
		for (register int j = 1; j < m + i; j++) {
			scanf("%d", &a[i][j]);
			p[i][j] = ++tj;
			add(p[i][j], p[i][j] + sum, 1, 0);
		}
	S = 0;
	E = 2 * sum + 1;
	SS = 2 * sum + 2;
	add(SS, S, m, 0);
	for (register int i = 1; i < m + n; i++)
		add(S, p[n][i], m, a[n][i]);
	for (register int i = 1; i <= m; i++)
		add(p[1][i] + sum, E, m, 0);
	for (register int i = 2; i <= n; i++) {
		for (register int j = 1; j < m + i; j++) {
			if (j < m + i - 1) add(p[i][j] + sum, p[i - 1][j], 1, a[i - 1][j]);
			if (j > 1) add(p[i][j] + sum, p[i - 1][j - 1], 1, a[i - 1][j - 1]);
		}
	}
	while (spfa())
		addflow();
	printf("%d\n", ans);
	tot = 1;
	ans = 0;
	memset(head, 0, sizeof(head));
	add(SS, S, m, 0);
	for (register int i = 1; i < m + n; i++)
		add(S, p[n][i], m, a[n][i]);
	for (register int i = 1; i <= m; i++)
		add(p[1][i] + sum, E, m, 0);
	for (register int i = 1; i <= n; i++) {
		for (register int j = 1; j < m + i; j++) {
			add(p[i][j], p[i][j] + sum, m, 0);
			if (i > 1 && j < m + i - 1) add(p[i][j] + sum, p[i - 1][j], 1, a[i - 1][j]);
			if (i > 1 && j > 1) add(p[i][j] + sum, p[i - 1][j - 1], 1, a[i - 1][j - 1]);
		}
	}
	while (spfa())
		addflow();
	printf("%d", ans);	
}