Flow Problem HDU

 Flow Problem HDU - 3549 

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. 

InputThe 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)OutputFor 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：网络流模版题，1到n的结点的最大的流
思路：用dinic

#include<stdio.h>
#include<iostream>
#include<map>
#include<string.h>
#include<queue>
#include<vector>
#include<math.h>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int dis[20];
int flow[500][500];
int n,m;
int bfs()
{
    memset(dis,-1,sizeof(dis));
    queue<int>Q;
    dis[1]=1;
    Q.push(1);
    while(!Q.empty())
    {
        int top=Q.front();
        Q.pop();
        for(int i=1;i<=n;i++)
        {
            if(flow[top][i]>0&&dis[i]<0)
            {
                dis[i]=dis[top]+1;
                Q.push(i);
            }
        }
    }
    if(dis[n]>0)return 1;
    return 0;
}
int dinic(int x,int k)
{
    if(x==n)
        return k;
    int y;
    for(int i=1;i<=n;i++)
    {
        if(flow[x][i]>0&&dis[i]==dis[x]+1&&(y=dinic(i,min(k,flow[x][i]))))
        {
            flow[x][i]-=y;
            flow[i][x]+=y;
            return y;
        }
    }
    return 0;
 
}
int main()
{
    int t;
    scanf("%d",&t);
    int cas=1;
    while(t--)
    {
        memset(flow,0,sizeof(flow));
        scanf("%d%d",&n,&m);
        int a,b,c;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            flow[a][b]+=c;
        }
        int ans=0;
        while(bfs())
        {
            //cout<<"++"<<endl;
            int res;
            while(res=dinic(1,inf))ans+=res;
        }
        printf("Case %d: %d
",cas++,ans);
    }
}

用sap加邻接矩阵，maze那里是叠加，不是覆盖

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e3+5;
const int  mod = 1e9+7;

int maze[maxn][maxn];
int gap[maxn],dis[maxn],pre[maxn],cur[maxn];

int sap(int start,int end ,int nodenum)
{
    memset(cur,0,sizeof cur);
    memset(dis,0,sizeof dis);
    memset(gap,0,sizeof gap);
    int u = pre[start] = start,maxflow = 0,aug = -1;
    gap[0] = nodenum;
    while(dis[start] < nodenum)
    {
        loop:
        for(int v = cur[u];v < nodenum; v++)
            if(maze[u][v] && dis[u] == dis[v] + 1){
                if(aug == -1 || aug > maze[u][v])
                    aug = maze[u][v];
                pre[v] = u;
                u = cur[u] = v;
                if(v == end)
                {
                    maxflow += aug;
                    for(u = pre[u]; v!=start;v=u,u=pre[u])
                    {
                        maze[u][v] -= aug;
                        maze[v][u] += aug;
                    }
                    aug = -1;
                }
                goto loop;
            }
        int mindis = nodenum - 1;
        for(int v = 0;v<nodenum;v++)
            if(maze[u][v] && mindis > dis[v])
            {
                cur[u] = v;
                mindis = dis[v];
            }
        if((--gap[dis[u]]) == 0)
            break;
        gap[dis[u] = mindis + 1]++;
        u = pre[u];

    }
    return maxflow;
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int ca = 1;ca <= t; ca++)
    {
        memset(maze,0,sizeof maze);
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            int a,b,l;
            scanf("%d%d%d",&a,&b,&l);
            maze[a-1][b-1] += l;
        }
        int res = sap(0,n-1,n);
        printf("Case %d: %d
",ca,res);
    }
}