HDU 3667 Transportation (最小费用最大流)

题意：

有N个城市M条道路，现在你要把K个物品从城市1转移到城市N，每条道路转移x个物品需要ai *x^2费用，并且每条道路都有转移物品个数的上限，问你转移这K个物品最小费用为多少，若不能转移，输出-1.

解题思路：

对于每条道路建边每个容量为1的费用分别为ai , 3*ai, 5*ai ... (x^2 - (x-1)^2) *ai ，因为如果流量为1，肯定是走ai这条路，如果流量为2，肯定是走ai和3ai，所以就转化成普通的最小费用最大流问题。

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;
#define pii pair<int,int>
#define mk_p make_pair

const int maxn = 100 + 5;
const int maxm = 500000 + 5;
const int INF = 2000000000;

struct Edge {
    int u, to, cap, cost, next;
    Edge(){}
    Edge(int _u, int _to, int _cap, int _cost, int _next) {
        u = _u; to = _to; cap = _cap; cost = _cost; next = _next;
    }
}edge[maxm<<1];

int head[maxn], E, dist[maxn], pp[maxn];
bool vis[maxn];

void init(int n) {
    for(int i = 0;i <= n; i++) head[i] = -1;
    E = 0;
}

void newedge(int u, int to, int cap, int cost) {
    edge[E] = Edge(u, to, cap, cost, head[u]);
    head[u] = E++;
    edge[E] = Edge(to, u, 0, -cost, head[to]);
    head[to] = E++;
}

int qu[333333];

bool SPFA(int s, int t, int n) {
    for(int i = 0;i <= n; i++) {
        vis[i] = 0; pp[i] = -1; dist[i] = INF;
    }
    vis[s] = true;
    int st = 0, ed = 0;
    qu[ed++] = s;
    dist[s] = 0;
    while(st < ed) {
        int u = qu[st++];
        vis[u] = false;
        for(int i = head[u];i != -1;i = edge[i].next) {
            int to = edge[i].to;
            int tmp = dist[u] + edge[i].cost;
            if(edge[i].cap && dist[to] > tmp) {
                dist[to] = tmp;
                pp[to] = i;
                if(!vis[to]) {
                    qu[ed++] =to;
                    vis[to] = true;
                }
            }
        }
    }
    if(dist[t] == INF)  return false;
    return true;
}

// 最小费用最大流
pii MCMF(int s, int t, int n) {
    int flow = 0;
    int mincost = 0;
    while(SPFA(s, t, n)) {
        int minflow = INF;
        for(int i = pp[t];i != -1;i = pp[edge[i].u]) if(minflow > edge[i].cap)
            minflow = edge[i].cap;
        flow += minflow;
        for(int i = pp[t];i != -1;i = pp[edge[i].u]) {
            edge[i].cap -= minflow;
            edge[i^1].cap += minflow;
        }
        mincost += dist[t]*minflow;
    }
    return mk_p(flow, mincost);
}

int n, m, k, u, to, a, c;
void solve() {
    init(n+1);
    // 建边
    for(int i = 0;i < m; i++) {
        scanf("%d%d%d%d", &u, &to, &a, &c);
        int top = min(k, c);
        for(int j = 1;j <= top; j++)
            newedge(u, to, 1, a*( j*j - (j-1)*(j-1)));
    }
    newedge(n, n+1, k, 0);
    pii ans = MCMF(1, n+1, n+1);
    if(ans.first < k)  puts("-1");
    else    printf("%d\n", ans.second);
}

int main() {
    while(scanf("%d%d%d", &n, &m, &k) != -1) {
        solve();
    }
    return 0;
}