HDU 1532 Drainage Ditches (最大网络流)

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 896    Accepted Submission(s): 414

Problem Description

Everytime it rains on Farmer John's fields, a pond forms over Bessie'sfavorite clover patch. This means that the clover is covered by waterfor awhile and takes quite a long time to regrow. Thus, Farmer John hasbuilt a set of drainage ditches so that Bessie's clover patch is nevercovered in water. Instead, the water is drained to a nearby stream.Being an ace engineer, Farmer John has also installed regulators at thebeginning of each ditch, so he can control at what rate water flowsinto that ditch.
Farmer John knows not only how many gallons ofwater each ditch can transport per minute but also the exact layout ofthe ditches, which feed out of the pond and into each other and streamin a potentially complex network.
Given all this information,determine the maximum rate at which water can be transported out of thepond and into the stream. For any given ditch, water flows in only onedirection, but there might be a way that water can flow in a circle.

 

 

Input

Theinput includes several cases. For each case, the first line containstwo space-separated integers, N (0 <= N <= 200) and M (2 <= M<= 200). N is the number of ditches that Farmer John has dug. M isthe number of intersections points for those ditches. Intersection 1 isthe pond. Intersection point M is the stream. Each of the following Nlines 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 theditch.

 

 

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

 

 

Sample Input

5 41 2 401 4 202 4 202 3 303 4 10

 

Sample Output

50

解题：

       最大网络流的模板题，从网上找了一个不错的模板，而有待理解。

#include <iostream>using namespace std;#define MAXN 205#define INF 2110000000#define MIN(x,y) (x<y?x:y)int map[MAXN][MAXN];int max_flow(int num,int map[][MAXN],int source,int sink)//参数含义：结点数量 网络 源点 汇点{int my_queue[MAXN],queue_first,queue_end;//数组做队列 实现BFS搜索路径 int pre[MAXN],min_flow[MAXN];//记录结点的父节点 当前路径中最小的一段的值，也即限制值int flow[MAXN][MAXN];//记录当前网络中的流int ans=0;//最终结果memset(flow,0,sizeof(flow));while(1)//一直循环，直到不存在增广路径{queue_first=0;//初始化队列queue_end=0;my_queue[queue_end++]=source;memset(pre,-1,sizeof(pre));min_flow[source]=INF;pre[source]=-2;//源点的父节点需特殊标示while(queue_first<queue_end)//BFS寻找增广路径{int temp=my_queue[queue_first++];//出队列for(int i=0;i<num;i++)//由结点temp往外扩展{if(pre[i]==-1&&flow[temp][i]<map[temp][i])//当结点i还未被探索到，并且还有可用流量{my_queue[queue_end++]=i;//加入队列pre[i]=temp;//标示父节点min_flow[i]=MIN(min_flow[temp],(map[temp][i]-flow[temp][i]));//求得min_flow}}if(pre[sink]!=-1)//sink的父节点不为初始值，说明BFS已经找到了一条路径{int k=sink;while(pre[k]>=0){flow[pre[k]][k]+=min_flow[sink];//将新的流量加入flowflow[k][pre[k]]=-flow[pre[k]][k];k=pre[k];}break;}}if(pre[sink]==-1) return ans;//不存在增广路径，返回else ans+=min_flow[sink];}}int main(){int m,n;while(cin>>n>>m){int a,b,cost;memset(map,0,sizeof(map));for(int i=0;i<n;i++){cin>>a>>b>>cost;map[a-1][b-1]+=cost;}cout<<max_flow(m,map,0,m-1)<<endl;}return 0;}