poj 1459 多源多匯最大流模板題 Dinic

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 29855		Accepted: 15447

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

題意： 給你n個點m條有向邊，有np個電站，nc個用戶，每一個電站都可以提供有限的電能，每個用戶都有一定對電的需求，每條邊都有最大的流通電流的容量，問這nc個用戶最多可以獲得多少的電能。

思路： 多源多匯最大流模板題，構造一個超級源點，超源指向所有的源點，正向容量爲該電站可以提供的電能，反向隨意（因爲廣搜的dep標記是不會讓他往超源走的），所有的匯點指向一個超級匯點，邊的容量爲原匯點所需的容量，然後直接Dinic（詳情請看我上一篇博客）。注意的是，以前我還沒注意優化，直到T了無數次後才發現在dfs的時候還可以繼續大大的優化，以往是每個dfs只找一條可行流，優化後就是每個點所有可行流都訪問並結合大大節約了時間，具體看代碼。

注意： 本題有點坑，那就是輸入的時候需要輸入括號和逗號，如果簡單隻用%c會出大問題，所以有很多種解決的辦法，詳情看代碼。

代碼

#include<stdio.h>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue> 
#define ll long long
#define qq printf("QAQ\n");
using namespace std;
const int maxn=1e5+5;
const int inf=0x3f3f3f3f;
const double e=exp(1.0);
const double pi=acos(-1);
struct Edge{
	int to,cap,rev;
};
vector<Edge>v[205];//101 爲超級源點 102爲超級匯點 
int dep[205],iter[205];
int n,np,nc,m;
bool bfs()
{
	memset(dep,0,sizeof dep);
	queue<int>q;
	q.push(101);
	dep[101]=1;
	while(!q.empty())
	{
		int now=q.front();
		q.pop();
		for(int i=0;i<v[now].size();i++)
		{
			if(v[now][i].cap>0&&dep[v[now][i].to]==0)
			{
				dep[v[now][i].to]=dep[now]+1;
				q.push(v[now][i].to);
			}
		}
	}
	return dep[102]>0;
}
int dfs(int s,int e,int flow)
{
	if(s==e)return flow;//flow爲當前可行弧的可行流 
	int f;
	int ff=0;	
	//for(int &i=iter[s];i<v[s].size();i++)2
	for(int i=0;i<v[s].size();i++)//1
	{
		//if(v[s][i].cap>0&&dep[s]==dep[v[s][i].to]-1) 2
		if(v[s][i].cap>0&&flow>ff&&dep[s]==dep[v[s][i].to]-1)//1
		{
			//f=dfs(v[s][i].to,e,min(flow,v[s][i].cap)); 2
			f=dfs(v[s][i].to,e,min(flow-ff,v[s][i].cap));// 1
			if(f>0){
				v[s][i].cap-=f;
				v[v[s][i].to][v[s][i].rev].cap+=f;
				ff+=f;// s點所有可能的流量 
				//return f;2
			}
		}
	}
	if(ff==0)dep[s]=0;//1
	return ff;//1  方法1 相當於把一個點上連接的點全部遍歷一遍後再返回 比法內容2要多點 效率稍稍比法2低(法1 110ms 法2 79ms) 但是理解起來非常容易
//	return 0; 2 //方法2 是用了一個 iter 數組來記錄 看了好久沒看懂  比不優化的代碼只多一點點 就多一個i取值和歸零
}
int Dinic()
{
	int max_flow=0,flow=0;
	while(bfs())
	{
		//memset(iter,0,sizeof iter); 2
		while(flow=dfs(101,102,inf))
		max_flow+=flow;
	}
	return max_flow;
}
int main()
{
	while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
	{
		int s,e,c;
		for(int i=0;i<n;i++)v[i].clear();
		v[101].clear();v[102].clear();
		char cc;
		for(int i=0;i<m;i++)
		{
		       //cin>>cc>>s>>cc>>e>>cc>>c; 用cin輸入輸出 不用管回車空格的問題 但是非常耗時間跑了差不多800ms 而scanf只用了100ms左右
			scanf(" (%d,%d)%d",&s,&e,&c);
			
			if(s==e)continue;
			v[s].push_back((Edge){e,c,v[e].size()});
			v[e].push_back((Edge){s,0,v[s].size()-1});
		}
		for(int i=0;i<np;i++)
		{			
			//cin>>cc>>e>>cc>>c;
			scanf(" (%d)%d",&e,&c);
			
			v[101].push_back((Edge){e,c,v[e].size()}); 
			v[e].push_back((Edge){101,inf,v[101].size()-1}); 
		}
		for(int i=0;i<nc;i++)
		{			
			//cin>>cc>>e>>cc>>c;				
			scanf(" (%d)%d",&e,&c);
			
			v[102].push_back((Edge){e,inf,v[e].size()}); 
			v[e].push_back((Edge){102,c,v[102].size()-1}); 
		}
		printf("%d\n",Dinic());		
	}
	return 0;
}