POJ 1459 Power Network (Dinic非递归 或 递归（强大））

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 20351		Accepted: 10698

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con. 

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6. 

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c_max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

Source

Southeastern Europe 2003

dinic非递归形式：

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int VM=150;
const int EM=20500;
const int INF=0x3f3f3f3f;

struct Edge{
    int u,v,nxt;
    int cap;
}edge[EM];

int n,np,nc,m,cnt,head[VM];
int src,des,dep[VM],que[VM];

void addedge(int cu,int cv,int cw){
    edge[cnt].u=cu;    edge[cnt].v=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
    edge[cnt].v=cv;    edge[cnt].v=cu;    edge[cnt].cap=0;    edge[cnt].nxt=head[cv];
    head[cv]=cnt++;
}
/*
int BFS(){
    int front=0,rear=0;
    memset(dep,-1,sizeof(dep));
    que[rear++]=src;
    dep[src]=0;
    while(front!=rear){
        int u=que[front++];
        front%=VM;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].v;
            if(edge[i].cap>0 && dep[v]==-1){
                dep[v]=dep[u]+1;
                que[rear++]=v;
                rear%=VM;
                if(v==des)      //优化1
                    return 1;
            }
        }
    }
    return 0;
}
*/          //Poj上用下面的BFS快些，HDU上用下面的快些
int BFS(){  
    queue<int> q;  
    while(!q.empty())  
        q.pop();  
    memset(dep,-1,sizeof(dep));  
    dep[src]=0;  
    q.push(src);  
    while(!q.empty()){  
        int u=q.front();  
        q.pop();  
        for(int i=head[u];i!=-1;i=edge[i].nxt){  
            int v=edge[i].v;  
            if(edge[i].cap>0 && dep[v]==-1){  
                dep[v]=dep[u]+1;  
                q.push(v);  
            }  
        }  
    }  
    if(dep[des]==-1)  
        return 0;  
    return 1;  
} 

int Dinic(){
    int ans=0;
    int stack[VM],next[VM]; ////stack[i为栈，存储当前增广路,  next[i]存储当前点的后继 跟head是一样的
    while(BFS()){
        memcpy(next,head,sizeof(head));
        int u=src, top=0;   //u为当前结点
        while(1){
            if(u==des){     //增广路已全部进栈
                int minx=INF, loc;
                for(int i=0;i<top;i++)  //找最小的增广跟并loc记录其在stack中位置
                    if(minx>edge[stack[i]].cap){    //以便退回该边继续DFS
                        minx=edge[stack[i]].cap;
                        loc=i;
                    }
                for(int i=0;i<top;i++){     //将增广路中的所有边修改
                    edge[stack[i]].cap-=minx;
                    edge[stack[i]^1].cap+=minx;
                }
                ans+=minx;
                top=loc;
                u=edge[stack[top]].u;   //当前结点修改为最小边的起点
            }
            for(int i=next[u];i!=-1;next[u]=i=edge[i].nxt)  //找到当前结点对应的下一条边
                if(edge[i].cap>0 && dep[edge[i].v]==dep[u]+1)   //不满足条件时，修改cur值
                    break;  //（去掉不合适的占）eg：1-->2 1-->3 1-->4 有边 但只有1-->4 这条边满足条件 就把1到2、3的边给去掉
            if(next[u]!=-1){    //当前结点的下一条边存在
                stack[top++]=next[u];   //把该边放入栈中
                u=edge[next[u]].v;      //再从下个点开始找
            }else{
                if(top==0)  //当前结点无未遍历的下一条边且栈空，DFS找不到下一条增广路
                    break;
                dep[u]=-1;  //当前结点不在增广路中，剔除该点
                u=edge[stack[--top]].u;     //退栈 回朔，继续查找
            }
        }
    }
    return ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    char str[30];
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m)){
        cnt=0;
        memset(head,-1,sizeof(head));
        src=n,  des=n+1;
        int u,v,w;
        while(m--){
            scanf("%s",str);
            sscanf(str,"(%d,%d)%d",&u,&v,&w);
            addedge(u,v,w);
        }
        while(np--){
            scanf("%s",str);
            sscanf(str,"(%d)%d",&v,&w);
            addedge(src,v,w);
        }
        while(nc--){
            scanf("%s",str);
            sscanf(str,"(%d)%d",&u,&w);
            addedge(u,des,w);
        }
        printf("%d\n",Dinic());
    }
    return 0;
}

 

 

2,Dinic 递归

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int VM=150;
const int EM=20500;
const int INF=0x3f3f3f3f;

struct Edge{
    int to,nxt;
    int cap;
}edge[EM];

int n,np,nc,m,cnt,head[VM];
int src,des,dep[VM];

void addedge(int cu,int cv,int cw){
    edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
    edge[cnt].to=cu;    edge[cnt].cap=0;    edge[cnt].nxt=head[cv];
    head[cv]=cnt++;
}

int BFS(){
    queue<int> q;
    while(!q.empty())
        q.pop();
    memset(dep,-1,sizeof(dep));
    dep[src]=0;
    q.push(src);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].to;
            if(edge[i].cap>0 && dep[v]==-1){
                dep[v]=dep[u]+1;
                q.push(v);
            }
        }
    }
    if(dep[des]==-1)
        return 0;
    return 1;
}

int DFS(int u,int minx){
    if(u==des)
        return minx;
    int tmp;
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        if(edge[i].cap>0 && dep[v]==dep[u]+1 && (tmp=DFS(v,min(edge[i].cap,minx)))){
            edge[i].cap-=tmp;
            edge[i^1].cap+=tmp;
            return tmp;
        }
    }
    dep[u]=-1;      //加了这条语句让原来TLE变得不再问题。。。。。。。Orz............
    return 0;
}

int Dinic(){
    int ans=0,tmp;
    while(BFS()){
        while(1){
            tmp=DFS(src,INF);
            if(tmp==0)
                break;
            ans+=tmp;
        }
    }
    return ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    char str[30];
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m)){
        cnt=0;
        memset(head,-1,sizeof(head));
        src=n,  des=n+1;
        int u,v,w;
        while(m--){
            scanf("%s",str);
            sscanf(str,"(%d,%d)%d",&u,&v,&w);
            addedge(u,v,w);
        }
        while(np--){
            scanf("%s",str);
            sscanf(str,"(%d)%d",&v,&w);
            addedge(src,v,w);
        }
        while(nc--){
            scanf("%s",str);
            sscanf(str,"(%d)%d",&u,&w);
            addedge(u,des,w);
        }
        printf("%d\n",Dinic());
    }
    return 0;
}