HDU_1532 && HDU_3549(最大流EK算法模板)

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12707    Accepted Submission(s): 6062

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
   
   
   
   

 
题意：从1开始到终点n，最多能通过多少的流量，就比如说1到2的流量是20，2到3的流量是10，那么到达终点3的流量为10，而1到2还能经过10的流量，2到3不能再通过。
分析：完全是照刘汝佳的紫书模板写的，EdmondsKarp算法，简称EK算法。
关于EK算法的详细解释我觉得这篇较清晰： http://www.wutianqi.com/?p=3107

 题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1532
代码清单：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 200 + 5;
const int INF = 1e9 + 5;

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 EdmondsKarp{
    int n,m;
    vector<Edge>edge; //边数的两倍
    vector<int>G[maxn]; //邻接表，G[i][j]表示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();
        edge.clear();
    }

    void AddEdge(int from,int to,int cap){
        edge.push_back(Edge(from,to,cap,0));
        edge.push_back(Edge(to,from,0,0)); //反向弧
        m=edge.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;
            while(!q.empty()) q.pop();
            q.push(s);
            a[s]=INF;
            while(!q.empty()){
                int x=q.front();q.pop();
                for(int i=0;i<G[x].size();i++){
                    Edge& e=edge[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]) return flow;
            for(int u=t;u!=s;u=edge[p[u]].from){
                edge[p[u]].flow+=a[t];
                edge[p[u]^1].flow-=a[t];
            }
            flow+=a[t];
        }
    }
};

int N,M;
int a,b,c;

void solve(){
    EdmondsKarp EK;
    EK.init(N);
    for(int i=0;i<N;i++){
        scanf("%d%d%d",&a,&b,&c);
        EK.AddEdge(a,b,c);
    }
    printf("%d\n",EK.Maxflow(1,M));

}

int main(){
    while(scanf("%d%d",&N,&M)!=EOF){
        solve();
    }return 0;
}

 

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11118    Accepted Submission(s): 5274

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

 

Input

The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

 

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

 

Sample Input

    
    
    
    

     
     
     
     2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     Case 1: 1
Case 2: 2
    
    
    
    

 

题意：最大流裸题。从1到n最大流量。
分析：同上题，最大流EK算法即可解。
题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=3549
代码清单：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 200 + 5;
const int INF = 1e9 + 5;

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 EdmondsKarp{
    int n,m;
    vector<Edge>edge; //边数的两倍
    vector<int>G[maxn]; //邻接表，G[i][j]表示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();
        edge.clear();
    }

    void AddEdge(int from,int to,int cap){
        edge.push_back(Edge(from,to,cap,0));
        edge.push_back(Edge(to,from,0,0));
        m=edge.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;
            while(!q.empty()) q.pop();
            q.push(s);
            a[s]=INF;
            while(!q.empty()){
                int x=q.front();q.pop();
                for(int i=0;i<G[x].size();i++){
                    Edge& e=edge[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]) return flow;
            for(int u=t;u!=s;u=edge[p[u]].from){
                edge[p[u]].flow+=a[t];
                edge[p[u]^1].flow-=a[t];
            }
            flow+=a[t];
        }
    }
};

int T;
int N,M;
int a,b,c;
int cases;

void solve(){
    scanf("%d%d",&N,&M);
    EdmondsKarp EK;
    EK.init(N);
    for(int i=0;i<M;i++){
        scanf("%d%d%d",&a,&b,&c);
        EK.AddEdge(a,b,c);
    }
    printf("Case %d: %d\n",++cases,EK.Maxflow(1,N));

}

int main(){
    cases=0;
    scanf("%d",&T);
    while(T--){
        solve();
    }return 0;
}