hdu3549 Flow Problem（dinic算法和ISAP算法）

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

23 21 2 12 3 13 31 2 12 3 11 3 1

 

Sample Output

Case 1: 1Case 2: 2

ISAP算法代码：

#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算法代码：

#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，原因就是写错了一个地方Orz。
k=dinic(i,min(sum,G[s][i])))!=0与<span style="font-family: 'Courier New';">k=dinic(i,min(sum,G[s][i]))!=0。。。</span>
<span style="font-family: 'Courier New';">因为赋值符号的优先级比不等于判断要低。。这个错误太隐蔽了。。Orz</span>