Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11855    Accepted Submission(s): 5631

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

ISAP演算法程式碼：

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#include<fstream>
using namespace std;
#define N 205
#define INF (1<<30)
#define min(x,y) (x<y?x:y)
int G[N][N],num[N],d[N],pre[N],n;
queue<int> q;
void bfs(){
    memset(d,-1,sizeof(d));
    memset(num,0,sizeof(num));
    d[n]=0;num[0]=1;q.push(n);int u;
    while(!q.empty()){
        u=q.front();q.pop();
        for(int i=1;i<=n;++i){
            if(d[i]==-1&&G[i][u]>0){
                d[i]=d[u]+1;q.push(i);
                ++num[d[i]];
            }
        }
    }
    //printf("d:");for(int i=1;i<=n;++i) printf("%d ",d[i]);printf("\n");
}
int augment(){
    int u=n,res=INF;
    while(u!=1){
        res=min(res,G[pre[u]][u]);
        u=pre[u];
    }u=n;
    while(u!=1){
        G[pre[u]][u]-=res;
        G[u][pre[u]]+=res;
        u=pre[u];
    }
    //printf("res=%d\n",res);
    return res;
}
int isap(){
    int ans=0,u=1,m;bfs();bool flag;
    while(d[u]<n){flag=true;
        if(u==n){ans+=augment();u=1;}
        for(int i=1;i<=n;++i){
            if(d[i]+1==d[u]&&G[u][i]>0){
                flag=false;pre[i]=u;u=i;break;
            }
        }//printf("u=%d\n",u);
        if(flag){m=n-1;
            for(int i=1;i<=n;++i){
                if(G[u][i]>0&&d[i]!=-1) m=min(m,d[i]);
            }
            if(--num[d[u]]==0) break;
            ++num[d[u]=m+1];
            if(u!=1) u=pre[u];
        }
    }
    return ans;
}
int main(){
    int m,u,v,w,Case=1,t;scanf("%d",&t);
    /*ifstream fin;fin.open("C:\\Users\\l\\Desktop\\Learning File\\Programing File\\check\\data.txt");
    ofstream fout;fout.open("C:\\Users\\l\\Desktop\\Learning File\\Programing File\\check\\my.txt");*/
    while(t--){
        scanf("%d%d",&n,&m);
        memset(G,0,sizeof(G));
        for(int i=0;i<m;++i){
            //fin>>u>>v>>w;
            scanf("%d%d%d",&u,&v,&w);
            G[u][v]+=w;
        }
        printf("Case %d: %d\n",Case++,isap());
        /*fout<<isap()<<endl;
        cout<<Case++<<" data run!"<<endl;*/
    }//fin.close();fout.close();
    return 0;
}
 

dinic演算法程式碼：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define N 20
#define INF (1<<30)
#define min(x,y) (x<y?x:y)
int G[N][N],d[N],n;
queue<int> q;
bool bfs(int s){
    memset(d,-1,sizeof(d));
    q.push(s);d[s]=0;int u;
    while(!q.empty()){
        u=q.front();q.pop();
        for(int i=1;i<=n;++i){
            if(d[i]==-1&&G[u][i]>0){d[i]=d[u]+1;q.push(i);}
        }
    }
    if(d[n]==-1) return false;
    else return true;
}
int dinic(int s,int sum){
    if(s==n) return sum;
    int k;
    for(int i=1;i<=n;++i){
        if(d[s]+1==d[i]&&G[s][i]>0&&(k=dinic(i,min(sum,G[s][i])))!=0){
            G[s][i]-=k;G[i][s]+=k;return k;
        }
    }
}
int main(){
    int t,m,u,v,w,Case=1,ans;scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        memset(G,0,sizeof(G));
        for(int i=0;i<m;++i){
            scanf("%d%d%d",&u,&v,&w);
            G[u][v]+=w;
        }ans=0;
        while(bfs(1)) ans+=dinic(1,INF);
        printf("Case %d: %d\n",Case++,ans);
    }
    return 0;
}
 

之前dinic老是WA，原因就是寫錯了一個地方Orz。

k=dinic(i,min(sum,G[s][i])))!=0與<span style="font-family: 'Courier New';">k=dinic(i,min(sum,G[s][i]))!=0。。。</span>

<span style="font-family: 'Courier New';">因為賦值符號的優先順序比不等於判斷要低。。這個錯誤太隱蔽了。。Orz</span>