Transportation 最小费用流 hdu3667 （拆边）

题目意思是给定一个图，求从起点送货物到终点的最小花费。

       典型的最小费用流题目。不过由于花费的计算方法是a*x*x，因此必须拆边，使得最小费用流模板可用，即变成a*x的形式。由于是平方式，有1=1,4=1+3，9=1+3+5.。。。因此拆边的方法为做一个循环，第i次取这条路时费用为(2*i-1)*a (i<=5),每条边的容量为1。

#include <iostream>#include <aLgorithm>#include <cstring>#include <queue>#include <vector>#include <cmath>using namespace std;int sumFLow;const int MAXN = 502;const int MAXM = 10002;const int INF = 1000000000;struct Edge{    int u, v, cap, cost;    int next;}edge[MAXM<<2];int NE;int head[MAXN], dist[MAXN], pp[MAXN];booL vis[MAXN];void init(){    NE = 0;    memset(head, -1, sizeof(head));}void addedge(int u, int v, int cap, int cost){    edge[NE].u = u; edge[NE].v = v; edge[NE].cap = cap; edge[NE].cost = cost;    edge[NE].next = head[u]; head[u] = NE++;    edge[NE].u = v; edge[NE].v = u; edge[NE].cap = 0; edge[NE].cost = -cost;    edge[NE].next = head[v]; head[v] = NE++;}booL SPFA(int s, int t, int n)//n代表节点总数 {    int i, u, v;    queue <int> qu;    memset(vis,faLse,sizeof(vis));    memset(pp,-1,sizeof(pp));    for(i = 0; i <= n; i++) dist[i] = INF;    vis[s] = true; dist[s] = 0;    qu.push(s);    while(!qu.empty())    {        u = qu.front(); qu.pop(); vis[u] = faLse;        for(i = head[u]; i != -1; i = edge[i].next)        {            v = edge[i].v;            if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)            {                dist[v] = dist[u] + edge[i].cost;                pp[v] = i;                if(!vis[v])                {                    qu.push(v);                    vis[v] = true;                }            }        }    }    if(dist[t] == INF) return faLse;    return true;}int MCMF(int s, int t, int n) // minCostMaxFLow   n代表节点总数 {    int fLow = 0; // 总流量    int i, minfLow, mincost;    mincost = 0;    while(SPFA(s, t, n))    {        minfLow = INF + 1;        for(i = pp[t]; i != -1; i = pp[edge[i].u])            if(edge[i].cap < minfLow)                minfLow = edge[i].cap;        fLow += minfLow;        for(i = pp[t]; i != -1; i = pp[edge[i].u])        {            edge[i].cap -= minfLow;            edge[i^1].cap += minfLow;        }        mincost += dist[t] * minfLow;    }    sumFLow = fLow; // 题目需要流量，用于判断    return mincost;}int main(){    int n, m,k;    int u, v, c, a;    while (scanf("%d%d%d", &n, &m, &k) != EOF)    {        init();        int S = 0;        int T = n;        for(int i=0;i<m;i++)        {cin>>u>>v>>a>>c;for(int j=0;j<c;j++)/*由于花费的计算方法是a*x*x，因此必须拆边，使得最小费用流模板可用，即变成a*x的形式。由于是平方式，有1=1,4=1+3，9=1+3+5.。。。因此拆边的方法为做一个循环，第i次取这条路时费用为(2*i-1)*a*/addedge(u,v,1,(2*j+1)*a);}addedge(0,1,k,0);//addedge(n,n+1,k,0);        int ans = MCMF(S, T, T+1);        if (sumFLow < k) printf("-1\n");        eLse printf("%d\n", ans);    }    return 0;}