poj 1459(网络流)

Posted 2020-07-18 AC菜鸟机

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 26688		Accepted: 13874

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)20
7 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

15
6

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.

题意：有a个站点，其中b个发电站,c个居民家，其余的都是中转站，给出d条边以及这条边的容量,以及发电站和居民家的容量,问居民家能够获得的最大电量。。

题解:学到了怎么读入这种恶心数据的方法。。cin可以处理换行，然后其余的就是构图问题了，建立超级源点和发电站连接，边的容量为发电站的容量，建立超级汇点和居民家连接，边的容量同理，然后求最大流。

#include <stdio.h>
#include <algorithm>
#include <queue>
#include <string.h>
#include <math.h>
#include <iostream>
using namespace std;
const int N = 205;
const int INF = 999999999;
struct Edge{
    int v,w,next;
}edge[N*N];
int head[N];
int level[N];
void addEdge(int u,int v,int w,int &k){
    edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++;
    edge[k].v = u,edge[k].w=0,edge[k].next=head[v],head[v]=k++;
}
int BFS(int src,int des){
    queue<int >q;
    memset(level,0,sizeof(level));
    level[src]=1;
    q.push(src);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        if(u==des) return 1;
        for(int k = head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v,w=edge[k].w;
            if(level[v]==0&&w!=0){
                level[v]=level[u]+1;
                q.push(v);
            }
        }
    }
    return -1;
}
int dfs(int u,int des,int increaseRoad){
    if(u==des) return increaseRoad;
    int ret=0;
    for(int k=head[u];k!=-1;k=edge[k].next){
        int v = edge[k].v,w=edge[k].w;
        if(level[v]==level[u]+1&&w!=0){
            int MIN = min(increaseRoad-ret,w);
            w = dfs(v,des,MIN);
            edge[k].w -=w;
            edge[k^1].w+=w;
            ret+=w;
            if(ret==increaseRoad) return ret;
        }
    }
    return ret;
}
int Dinic(int src,int des){
    int ans = 0;
    while(BFS(src,des)!=-1) ans+=dfs(src,des,INF);
    return ans;
}

int main(){
    int a,b,c,d;
    while(cin>>a>>b>>c>>d){
        memset(head,-1,sizeof(head));
        int u,v,w,tot=0;
        char temp;
        for(int i=0;i<d;i++){
            cin>>temp>>u>>temp>>v>>temp>>w;
            if(u==v)continue;
            addEdge(u+1,v+1,w,tot);
        }
        for(int i=0;i<b;i++){
            cin>>temp>>u>>temp>>w;
            addEdge(0,u+1,w,tot);
        }
        for(int i=0;i<c;i++){
            cin>>temp>>u>>temp>>w;
            addEdge(u+1,a+1,w,tot);
        }
        printf("%d\n",Dinic(0,a+1));
    }
}