网络流裸题
http://acm.nefu.edu.cn/JudgeOnline/problemshow.php?problem_id=473
description |
Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. Note however, that there can be more than one ditch between two intersections.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
|
input |
Input file contains multiple test cases. In a test case: Line 1: Two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Line 2..N+1: Each of N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch. |
output |
For each case,One line with a single integer, the maximum rate at which water may emptied from the pond. |
sample_input |
5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10 |
sample_output |
50 |
解题思路：这题就是普通的最大流，明显的给出最大流模型，通常我们都是在纸上画出最大流的模型，然后根据模型建图，并计算规模，这题题目给出的就是最大流的模型，所以直接套模板就行了。
#include <stdio.h>
#include <iostream>
using namespace std;
const int oo=1e9;
const int mm=111111;
const int mn=999;
int node ,scr,dest,edge;
int ver[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node,int _scr,int _dest)
{
node=_node,scr=_scr,dest=_dest;
for(int i=0; i<node; ++i)
head[i]=-1;
edge=0;
}
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++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for(i=0; i<node; i++)
dis[i]=-1;
dis[q[r++]=scr]=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)
{
dis[q[r++]=v]=dis[u]+1;
if(v==dest)
return 1;
}
}
}
return 0;
}
int Dinic_dfs(int u,int exp)
{
if(u==dest)
return exp;
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;
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(scr,oo))
ret+=delta;
}
return ret;
}
int main()
{
int n,m,u,v,c;
while(~scanf("%d%d",&n,&m))
{
prepare(m+1,1,m);
while(n--)
{
scanf("%d%d%d",&u,&v,&c);
addedge(u,v,c);
}
printf("%d\n",Dinic_flow());
}
return 0;
}