HDU 3549 Flow Problem

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Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4875    Accepted Submission(s): 2282

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
   

 

Author

HyperHexagon

 

Source

HyperHexagon's Summer Gift (Original tasks)

 

Recommend

zhengfeng

求最大流问题，直接套模板。

此为EK算法，耗时比较长，在此之后有sap算法。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define INF 9999999
int n,m;
int map[300][300],pre[300],flow[300][300],p[300],a[300];
int EK(int s,int t)
{
    int sum=0;
    queue<int>q;
    memset(flow,0,sizeof(flow));
    for(;;)
    {
        memset(a,0,sizeof(a));//记录残量
        a[s]=INF;
        q.push(s);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int i=2; i<=n; i++)
                if(!a[i]&&map[u][i]>flow[u][i])
                {
                    p[i]=u;//记录i的父亲节的是u
                    q.push(i);
                    a[i]=a[u]<map[u][i]-flow[u][i]?a[u]:map[u][i]-flow[u][i];
                }
        }
        if(!a[t])break;//如果残量是0的话，就找到最大流
        for(int i=t; i!=s; i=p[i])//每条路加上最小残量
        {
            flow[p[i]][i]+=a[t];
            flow[i][p[i]]-=a[t];
        }
        sum+=a[t];//记录流量
    }
    return sum;
}
int main()
{
    int a,b,c,max,t;
    scanf("%d",&t);
    for(int textcase=1;textcase<=t;textcase++)
    {
         scanf("%d%d",&n,&m);
       {
        memset(map,0,sizeof(map));
        memset(p,0,sizeof(p));

        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            map[a][b]+=c;
        }
        max=EK(1,n);
        printf("Case %d: %d\n",textcase,max);
        }
    }
    return 0;
}

SAP算法是一种高效算法，如若不明白其中的道理，可参考这个： SAP参考 。

鄙人的挫折码。

#include<stdio.h>
#include<string.h>
#define inf 9999999
#define MIN(a,b) a>b?b:a;
struct E
{
    int to,val,next;
}edg[500];
int map[16][16],dis[16],gap[16],list[16],node;
int sourse,sink,vs;
void addedge(int from,int to,int val)
{
    edg[node].to=to;
    edg[node].val=val;
    edg[node].next=list[from];
    list[from]=node++;
    edg[node].to=from;
    edg[node].val=0;
    edg[node].next=list[to];
    list[to]=node++;
}
int dfs(int src,int aug)
{
    if(src==sink)return aug;
    int flow=0,MIN_d=vs-1;
    for(int j=list[src];j!=-1;j=edg[j].next)
        if(edg[j].val)
        {
           if(dis[src]==dis[edg[j].to]+1)
           {
               int minn=MIN(aug-flow,edg[j].val);
               int t=dfs(edg[j].to,minn);
               edg[j].val-=t;
               edg[j^1].val+=t;
               flow+=t;
               if(dis[sourse]>=vs)return flow;
               if(aug==flow)break;
           }
           MIN_d=MIN(MIN_d,dis[edg[j].to]);
        }
        if(!flow)
        {
            if(!(--gap[dis[src]]))dis[sourse]=vs;
            ++gap[dis[src]=MIN_d+1];
        }
        return flow;
}
int sap(int src,int ed)
{
    int ans=0;
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    gap[0]=vs=ed;
    sourse=src;
    sink=ed;
    while(dis[sourse]<vs)
    ans+=dfs(sourse,inf);
    return ans;
}
int main()
{
    int cas;
    scanf("%d",&cas);
    for(int k=1;k<=cas;k++)
    {
        memset(map,0,sizeof(map));
        memset(list,-1,sizeof(list));
        node=0;
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            int a,b,u;
            scanf("%d%d%d",&a,&b,&u);
            map[a][b]+=u;
        }
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        if(map[i][j])
        {
            addedge(i,j,map[i][j]);
        }
        int anss=sap(1,n);
        printf("Case %d: %d\n",k,anss);
    }
}