题目链接：https://vjudge.net/problem/HDU-3667
Transportation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3083 Accepted Submission(s): 1341
Problem DescriptionThere are N cities, and M directed roads connecting them. Now you want to transport K units of goods from city 1 to city N. There are many robbers on the road, so you must be very careful. The more goods you carry, the more dangerous it is. To be more specific, for each road i, there is a coefficient ai. If you want to carry x units of goods along this road, you should pay ai * x2 dollars to hire guards to protect your goods. And what’s worse, for each road i, there is an upper bound Ci, which means that you cannot transport more than Ci units of goods along this road. Please note you can only carry integral unit of goods along each road.
You should find out the minimum cost to transport all the goods safely. InputThere are several test cases. The first line of each case contains three integers, N, M and K. (1 <= N <= 100, 1 <= M <= 5000, 0 <= K <= 100). Then M lines followed, each contains four integers (ui, vi, ai, Ci), indicating there is a directed road from city ui to vi, whose coefficient is ai and upper bound is Ci. (1 <= ui, vi <= N, 0 < ai <= 100, Ci <= 5) OutputOutput one line for each test case, indicating the minimum cost. If it is impossible to transport all the K units of goods, output -1.
Sample Input2 1 2
1 2 1 2
2 1 2
1 2 1 1
2 2 2
1 2 1 2
1 2 2 2 Sample Output4
-1
3 Source2010 Asia Regional Harbin Recommendlcy
题解：
费用与流量平方成正比。详情在《训练指南》P366 。主要方法是拆边。
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+;
const int MAXM = 1e5+;
const int MAXN = 1e2+; struct Edge
{
int to, next, cap, flow, cost;
}edge[MAXM];
int tot, head[MAXN];
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N; void init(int n)
{
N = n;
tot = ;
memset(head, -, sizeof(head));
} void add(int u, int v, int cap, int cost)
{
edge[tot].to = v; edge[tot].cap = cap; edge[tot].cost = cost;
edge[tot].flow = ; edge[tot].next = head[u]; head[u] = tot++;
edge[tot].to = u; edge[tot].cap = ; edge[tot].cost = -cost;
edge[tot].flow = ; edge[tot].next = head[v]; head[v] = tot++;
} bool spfa(int s, int t)
{
queue<int>q;
for(int i = ; i<N; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -;
} dis[s] = ;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i!=-; 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]==-) return false;
return true;
} int minCostMaxFlow(int s, int t, int &cost)
{
int flow = ;
cost = ;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t]; i!=-; i = pre[edge[i^].to])
{
if(Min>edge[i].cap-edge[i].flow)
Min = edge[i].cap-edge[i].flow;
}
for(int i = pre[t]; i!=-; i = pre[edge[i^].to])
{
edge[i].flow += Min;
edge[i^].flow -= Min;
cost += edge[i].cost*Min;
}
flow += Min;
}
return flow;
} int main()
{
int n, m, K;
while(scanf("%d%d%d", &n, &m, &K)!=EOF)
{
init(n+);
for(int i = ; i<=m; i++)
{
int u, v, c, a;
scanf("%d%d%d%d", &u, &v, &a, &c);
for(int j = ; j<=c; j++) //拆边
add(u, v, , (j*-)*a); //拆成费用为1 3 5 7 9……的边，每条边的容量为1
}
add(, , K, ); int min_cost;
int start = , end = n;
int max_flow = minCostMaxFlow(start, end, min_cost); if(max_flow<K) printf("-1\n");
else printf("%d\n", min_cost);
}
}
