Drainage Ditches
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.
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.
InputThe 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.
OutputFor 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 10
Sample Output
50 最大流的模板 考虑重边
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<cstdlib>
6 #include<string.h>
7 #include<set>
8 #include<vector>
9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<cmath>
13 typedef long long ll;
14 typedef unsigned long long LL;
15 using namespace std;
16 const double PI=acos(-1.0);
17 const double eps=0.0000000001;
18 const int INF=1e9;
19 const int N=1000+100;
20 int mp[N][N];
21 int vis[N];
22 int pre[N];
23 int m,n;
24 int BFS(int s,int t){
25 queue<int>q;
26 memset(pre,-1,sizeof(pre));
27 memset(vis,0,sizeof(vis));
28 pre[s]=0;
29 vis[s]=1;
30 q.push(s);
31 while(!q.empty()){
32 int p=q.front();
33 q.pop();
34 for(int i=1;i<=n;i++){
35 if(mp[p][i]>0&&vis[i]==0){
36 pre[i]=p;
37 vis[i]=1;
38 if(i==t)return 1;
39 q.push(i);
40 }
41 }
42 }
43 return false;
44 }
45 int EK(int s,int t){
46 int flow=0;
47 while(BFS(s,t)){
48 //BFS(s,t);
49 int dis=INF;
50 for(int i=t;i!=s;i=pre[i])
51 dis=min(mp[pre[i]][i],dis);
52 for(int i=t;i!=s;i=pre[i]){
53 mp[pre[i]][i]=mp[pre[i]][i]-dis;
54 mp[i][pre[i]]=mp[i][pre[i]]+dis;
55 }
56 flow=flow+dis;
57 }
58 return flow;
59 }
60 int main(){
61 while(scanf("%d%d",&n,&m)!=EOF){
62 memset(mp,0,sizeof(mp));
63 for(int i=0;i<n;i++){
64 int u,v,w;
65 scanf("%d%d%d",&u,&v,&w);
66 mp[u][v]=mp[u][v]+w;
67 }
68 int ans=EK(1,m);
69 cout<<ans<<endl;
70 }
71
72 }