poj1459（最大流）

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p _max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l _max(u,v) of power delivered by u to v. Let Con=Σ _uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con. 

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p _max(u)=y. The label x/y of consumer u shows that c(u)=x and c _max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l _max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6. 

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l _max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p _max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c _max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)207 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

156

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

理解题目意思建边~！

    输入分别为m个点，a个发电站，b个用户，n条边；接下去是n条边的信息(u,v)cost，cost表示边(u,v)的最大流量；a个发电站的信息(u)cost，cost表示发电站u能提供的最大流量；b个用户的信息(v)cost，cost表示每个用户v能接受的最大流量。

然后呢这题的输入也不需要像别人说的sscanf那样，直接scanf也是能搞定的！

下面是代码，具体建边看代码理解~~：

#include<stdio.h>#include<iostream>using   namespace std;const   int oo=1e9; const  int mm=111111; const  int mn=999; int node,src,dest,edge; int ver[mm],flow[mm],next[mm]; int head[mn],work[mn],dis[mn],q[mn];void prepare(int _node, int _src,int _dest){ node=_node,src=_src,dest=_dest; for(int i=0;i<node;++i)head[i]=-1; edge=0; } /**增加一条 u 到 v 容量为 c 的边*/ void addedge( int u,  int v,  int c) { ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++; ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++; } /**广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束*/ bool Dinic_bfs() { int i,u,v,l,r=0; for(i=0;i<node;++i)dis[i]=-1; dis[q[r++]=src]=0; for(l=0;l<r;++l) for(i=head[u=q[l]];i>=0;i=next[i]) if(flow[i]&&dis[v=ver[i]]<0) {/**这条边必须有剩余容量*/ dis[q[r++]=v]=dis[u]+1; if(v==dest)  return 1; } return 0; } /**寻找可行流的增广路算法，按节点的距离来找，加快速度*/ int Dinic_dfs(  int u, int exp) { if(u==dest)  return exp; /**work 是临时链表头，这里用 i 引用它，这样寻找过的边不再寻找*/ for(  int &i=work[u],v,tmp;i>=0;i=next[i]) if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0) { flow[i]-=tmp; flow[i^1]+=tmp; /**正反向边容量改变*/ return tmp;} return 0; }int Dinic_flow() { int i,ret=0,delta; while(Dinic_bfs()) { for(i=0;i<node;++i)work[i]=head[i]; while(delta=Dinic_dfs(src,oo))ret+=delta; } return ret; }int main(){    int n,m,u,v,c,np,nc;    while(~scanf("%d%d%d%d",&n,&np,&nc,&m))    {        char ch;        prepare(n+3,0,n+1);        while(m--)        {            scanf(" %c%d%c%d%c%d",&ch,&u,&ch,&v,&ch,&c);            addedge(u+1,v+1,c);        }        while(np--)        {            scanf(" %c%d%c%d",&ch,&u,&ch,&c);            addedge(0,u+1,c);        }        while(nc--)        {            scanf(" %c%d%c%d",&ch,&u,&ch,&c);            addedge(u+1,n+1,c);        }        printf("%d\n",Dinic_flow());    }    return 0;}