最小费用最大流 hdu 3667

来源：互联网发布：阿玛拉王国捏脸数据编辑：程序博客网时间：2024/05/29 13:20
/*学了好几天了，终于生成了模板一直以来有一个问题没有克服，今天终于想明白了，那就是为什么要加反向流量呢？那是因为如果你在加流量的时候如果加多了，，则要溜回去。。。。*/#include<iostream>#include<queue>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn =  100000;const int inf = 0x777777;struct {    int head;}H[maxn];struct {    int u,v,next,cap,cost;}E[10*maxn];int top;void add(int u,int v,int cap,int cost) {    E[top].v = v;E[top].cap = cap;    E[top].cost = cost;    E[top].next = H[u].head;    H[u].head = top++;}int d[maxn];int p[maxn];int vis[maxn];int path[maxn];int max_flow,ans;bool spfa(int s,int t,int n){    for(int i=1;i<=n;i++)        d[i]  =  inf;        d[s] = 0;        memset(vis,0,sizeof(vis));        queue <int> q;        q.push(s);vis[s] = 1;        while(!q.empty())        {            int x = q.front();q.pop();vis[x] = 0; //cout<<x<<endl;            for(int i=H[x].head;i!=-1;i = E[i].next)            {                if(E[i].cap > 0 && d[E[i].v] > d[x]+E[i].cost)                {                    d[E[i].v] = d[x]+E[i].cost;                    path[E[i].v] = i;   //这个地方很是关键，记录当前边的编号                    p[E[i].v] = x;                    if(!vis[E[i].v])                    {                        vis[E[i].v] = 1;                        q.push(E[i].v);                    }                }            }        }                if(d[t] == inf) return false;        int u = t,sum = inf;        while (u!=s) {             sum = min(E[path[u]].cap,sum);            u = p[u];        }        u = t;        max_flow += sum;        while(u != s) {     // 更新正反流量            E[path[u]].cap -= sum;            E[path[u]^1].cap += sum;            ans += E[path[u]].cost * sum;            u = p[u];        }        return true;}void init(){    memset(H,-1,sizeof(H));    memset(E,-1,sizeof(E));    top = 0;    ans = 0;    max_flow = 0;}int main(){    int n,m,k;    int u,v,cost,cap;    while(~scanf("%d%d%d",&n,&m,&k))    {        init();        while(m--)        {            scanf("%d%d%d%d",&u,&v,&cost,&cap);            for(int i=1;i<=cap;i++)            {                add(u,v,1,cost*(2*i-1));                add(v,u,0,-cost*(2*i-1));            }        }        add(0,1,k,0);        add(1,0,0,0);        while(spfa(0,n,n));        if(max_flow<k)            printf("-1\n");        else            printf("%d\n",ans);    }}
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