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
u
c(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.
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
The meaning of the question: Given n points and m directed edges, there are np power stations and nc users. Each power station can provide a limited amount of electricity. Each user has a certain demand for electricity, and each edge has a maximum The capacity of the circulating current, ask the nc users how much power can be obtained at most.
Idea: Multi-source and multi-sink maximum flow template problem, construct a super source point, super source points to all source points, the forward capacity is the electric energy that the power station can provide, and the reverse direction is optional (because the dep mark of Guangsou will not allow He went to the super source), all the sinks point to a super sink, the capacity of the edge is the capacity required by the original sink, and then directly Dinic (for details, please refer to my last blog). It should be noted that I didn’t pay attention to optimization before. It was not until after T that I found that I could continue to optimize greatly in dfs. In the past, only one feasible flow was found in each dfs, and after optimization, all feasible flows at each point were found. Both access and combination save a lot of time, see the code for details.
Note: This question is a bit pitted, that is, you need to enter parentheses and commas when inputting. If you simply use %c, there will be a big problem, so there are many solutions, see the code for details.
code#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 is the super source point 102 is the super sink
int dep[205], path[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为当前可行弧的可行流
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点所有可能的流量
//return f;2
}
}
}
if(ff==0)dep[s]=0;//1
return ff;//1 方法1 相当于把一个点上连接的点全部遍历一遍后再返回 比法内容2要多点 效率稍稍比法2低(法1 110ms 法2 79ms) 但是理解起来非常容易
// return 0; 2 //方法2 是用了一个 iter 数组来记录 看了好久没看懂 比不优化的代码只多一点点 就多一个i取值和归零
}
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输入输出 不用管回车空格的问题 但是非常耗时间跑了差不多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;
}