Drainage Ditches |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 45 Accepted Submission(s): 38 |
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Problem 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. |
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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. |
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Output For each case, output a single integer, the maximum rate at which water may emptied from the pond. |
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Sample Input 5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10 |
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Sample Output 50 |
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Source USACO 93 |
题意：
裸的最大流
代码：
//紫书366页。模板。点的编号从0开始。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=202,inf=0x7fffffff;
struct edge{
int from,to,cap,flow;
edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Edmonds_Karp{
int n,m;
vector<edge>edges;//边数的两倍
vector<int>g[maxn];//邻接表，g[i][j]表示节点i的第j条边在e数组中的序号
int a[maxn];//当起点到i的可改进量
int p[maxn];//最短路树上p的入弧编号
void init(int n){
for(int i=0;i<n;i++) g[i].clear();
edges.clear();
}
void addedge(int from,int to,int cap){
edges.push_back(edge(from,to,cap,0));
edges.push_back(edge(to,from,0,0));//反向弧
m=edges.size();
g[from].push_back(m-2);
g[to].push_back(m-1);
}
int Maxflow(int s,int t){
int flow=0;
for(;;){
memset(a,0,sizeof(a));
queue<int>q;
q.push(s);
a[s]=inf;
while(!q.empty()){
int x=q.front();q.pop();
for(int i=0;i<(int)g[x].size();i++){
edge&e=edges[g[x][i]];
if(!a[e.to]&&e.cap>e.flow){
p[e.to]=g[x][i];
a[e.to]=min(a[x],e.cap-e.flow);
q.push(e.to);
}
}
if(a[t]) break;
}
if(!a[t]) break;
for(int u=t;u!=s;u=edges[p[u]].from){
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
}
flow+=a[t];
}
return flow;
}
}EK;
int main()
{
int n,m,a,b,c;
while(scanf("%d%d",&n,&m)==2){
EK.init(m);
for(int i=0;i<n;i++){
scanf("%d%d%d",&a,&b,&c);
a--;b--;
EK.addedge(a,b,c);
}
printf("%d\n",EK.Maxflow(0,m-1));
}
return 0;
}