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.
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 The input includes several cases. For each case, the first line contains 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. Each of the following 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, output 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 10Sample Output
50
1.Ford--Fulkerson算法
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int M = 1e5 + 10;
//表示邊的結構體(終點,容量,反向邊)
struct node
{
int to;ll cap;
int rev;
};
int n,m;
vector<node>no[1000];
bool vis[1000];
//向圖增加一條從s到t容量爲cap的邊
void init(int from,int to,ll cap)
{
no[from].push_back((node){to,cap,no[to].size()});
no[to].push_back((node){from,0,no[from].size()-1});
}
ll dfs(int v,int t,ll f)
{
if(v==t)
return f;
vis[v]=1;
for(int i=0;i<no[v].size();i++)
{
node &e=no[v][i];
if(!vis[e.to]&&e.cap>0)
{
ll d=dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
no[e.to][e.rev].cap+=d;
return d;
}
}
}
return 0;
}
//求解S到T的最大流
ll max_flow(int s,int t)
{
ll ff=0;
for(;;)
{
memset(vis,0,sizeof(vis));
ll f=dfs(s,t,inf);
if(f==0)
return ff;
ff+=f;
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(no,0,sizeof(no));
int fr,to; ll c;
for(int i=0;i<n;i++)
{
scanf("%d%d%lld",&fr,&to,&c);
init(fr,to,c);
}
ll ans=max_flow(1,m);
printf("%lld\n",ans);
}
return 0;
}