hdu 1532(poj 1273) Drainage Ditches (网络流·最大流)

hdu 1532(poj 1273) Drainage Ditches (网络流·最大流)

题目：​​Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12515    Accepted Submission(s): 5968

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.  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 10

Sample Output

50

关键点：有向图，可重边（最大通量累加），小数据量（EK算法）。

#include #include #include #include using namespace std;const int N=210,INF=0x3f3f3f3f;int n,m;int map[N][N],p[N];bool flag[N];bool EK_bfs(int start,int end){ queue q; memset(flag,0,sizeof(flag)); memset(p,-1,sizeof(p)); q.push(start); flag[start]=1; while(!q.empty()){ int e=q.front(); if(e==end) return 1; q.pop(); for(int i=1;i<=m;i++){ if(map[e][i]&&!flag[i]){ flag[i]=1; p[i]=e; q.push(i); } } } return 0;}int EK_maxflow(int start,int end){ int u,ans=0,temp; while(EK_bfs(start,end)){ temp=INF; u=end; while(p[u]!=-1){ temp=min(temp,map[p[u]][u]); u=p[u]; } ans+=temp; u=end; while(p[u]!=-1){ map[p[u]][u]-=temp; map[u][p[u]]+=temp; u=p[u]; } } return ans;}int main(){ //freopen("cin.txt","r",stdin); while(cin>>n>>m){ int a,b,c; memset(map,0,sizeof(map)); for(int i=0;i

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。