ZOJ 3231 - Apple Transportation（上下界費用流）

題目：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3231

題意：

n個點 有點權和邊權的樹，要使得樹上的點權盡量平均，求出最小花費。

思路：

最終的狀态是 每個點的點權是 avg 或者avg+1，則每個點都有一個容量下界avg，每一個連一條容量為avg的邊到彙點t。而容量上界為avg+1，則每一個點還要連一條容量為1的邊 連到虛拟彙點，虛拟彙點連到彙點的容量為 sum%n。

AC.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>
#include <string>

#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int MAXN = 1000;
const int MAXM = 50000;
const int INF = 0x3f3f3f3f;

typedef long long ll;

struct Edge
{
	int to, next, cap, flow, cost;
} edge[MAXM];

int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;//節點總個數，節點編号從0~N-1

void init(int n)
{
	N = n;
	tol = 0;
	memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap, int cost)
{
	edge[tol].to = v;
	edge[tol].cap = cap;
	edge[tol].cost = cost;
	edge[tol].flow = 0;
	edge[tol].next = head[u];
	head[u] = tol++;
	edge[tol].to = u;
	edge[tol].cap = 0;
	edge[tol].cost = -cost;
	edge[tol].flow = 0;
	edge[tol].next = head[v];
	head[v] = tol++;
}

bool spfa(int s, int t)
{
	queue<int>q;
	for (int i = 0; i < N; i++)
	{
		dis[i] = INF;
		vis[i] = false;
		pre[i] = -1;
	}
	dis[s] = 0;
	vis[s] = true;
	q.push(s);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		vis[u] = false;
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			if (edge[i].cap > edge[i].flow &&
			        dis[v] > dis[u] + edge[i].cost )
			{
				dis[v] = dis[u] + edge[i].cost;
				pre[v] = i;
				if (!vis[v])
				{
					vis[v] = true;
					q.push(v);
				}
			}
		}
	}
	if (pre[t] == -1) return false;
	else return true;
}

//傳回的是最大流，cost存的是最小費用
int flow(int s, int t, int &cost)
{
	int flow = 0;
	cost = 0;
	while (spfa(s, t))
	{
		int Min = INF;
		for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
		{
			if (Min > edge[i].cap - edge[i].flow)
				Min = edge[i].cap - edge[i].flow;
		}
		for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
		{
			edge[i].flow += Min;
			edge[i ^ 1].flow -= Min;
			cost += edge[i].cost * Min;
		}
		flow += Min;
	}
	return flow;
}

int val[MAXN];
int main()
{
   // freopen("in", "r", stdin);
    int n;
    while(~scanf("%d", &n)) {
        int s = 0, tt = n+1, t = n+2, sum = 0;
        int u, v, w;
        init(t+1);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &val[i]);
            sum += val[i];
        }
        int avg = sum/n;
        for(int i = 1; i < n; ++i) {
            scanf("%d%d%d", &u, &v, &w);
            u++; v++;
            addedge(u, v, INF, w);
            addedge(v, u, INF, w);
        }

        for(int i = 1; i <= n; ++i) {
            addedge(s, i, val[i], 0);
            addedge(i, t, avg, 0);
            addedge(i, tt, 1, 0);
        }
        addedge(tt, t, sum%n, 0);

        int cost = 0;
        int ans = flow(s, t, cost);
        printf("%d\n", cost);
    }
    return 0;
}