标签：acm edmondkarp poj
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
Output
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define maxn 100+5
#define inf 0x7fffffff
int n,np,nc,m;
int s,t,edge[maxn][maxn],pre[maxn];
int EK()
{
int minflow,maxflow = 0;
while(1)
{
memset(pre, 0, sizeof(pre));
minflow = inf;
queue<int> q;
q.push(s);
while(!q.empty()) //BFS寻找增广路
{
int u = q.front();
q.pop();
for(int i = 0;i < n+2;i ++)
{
if(edge[u][i] > 0 && pre[i] == 0 )
{
pre[i] = u;
q.push(i);
}
}
}
//cout << "here" << endl;
if(pre[t] == 0) break;
for(int i = t; i != s;i = pre[i])
minflow = min(minflow, edge[pre[i]][i]);
for(int i = t; i != s;i = pre[i])
{
edge[pre[i]][i] -= minflow;
edge[i][pre[i]] += minflow;
}
maxflow += minflow;
//cout << maxflow <<endl;
}
return maxflow;
}
int main()
{
char a,b,c;
int from,to,cost;
while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF)
{
s = n ;
t = n + 1;
for(int i = 1; i <= m;i ++)
{
cin >> a >> from >> b >> to >> c >> cost;
edge[from][to] = cost;
//cout <<from <<" "<<to<<" "<<cost<<endl;
}
for(int i = 0; i < np;i ++)
{
cin >> a >> to >> b >> cost;
edge[s][to] = cost;
//cout <<s<<" "<<to <<" "<<cost <<endl;
}
for(int i = 0; i < nc;i ++)
{
cin >>a>>from>>b>>cost;
edge[from][t] = cost;
//cout <<from <<" "<<t<<" "<<cost<<endl;
}
int ans = EK();
printf("%d\n", ans);
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
标签：acm edmondkarp poj
原文地址：http://blog.csdn.net/mowenwen_/article/details/47838231
