HDU1532最大流 Edmonds-Karp,Dinic算法模板

時間 2019-11-21

標籤 hdu1532 hdu 最大 edmonds karp dinic 算法模板简体版

原文原文鏈接

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 45 Accepted Submission(s): 38

Problem Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

Sample Output

Source

USACO 93

題意：ios

裸的最大流算法

代碼：數組

//Edmonds-Karp算法，紫書366頁。模板。點的編號從0開始。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=202,inf=0x7fffffff;
struct edge{
    int from,to,cap,flow;
    edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Edmonds_Karp{
    int n,m;
    vector<edge>edges;//邊數的兩倍
    vector<int>g[maxn];//鄰接表，g[i][j]表示節點i的第j條邊在e數組中的序號
    int a[maxn];//當起點到i的可改進量
    int p[maxn];//最短路樹上p的入弧編號
    void init(int n){
        for(int i=0;i<n;i++) g[i].clear();
        edges.clear();
    }
    void addedge(int from,int to,int cap){
        edges.push_back(edge(from,to,cap,0));
        edges.push_back(edge(to,from,0,0));//反向弧
        m=edges.size();
        g[from].push_back(m-2);
        g[to].push_back(m-1);
    }
    int Maxflow(int s,int t){
        int flow=0;
        for(;;){
            memset(a,0,sizeof(a));
            queue<int>q;
            q.push(s);
            a[s]=inf;
            while(!q.empty()){
                int x=q.front();q.pop();
                for(int i=0;i<(int)g[x].size();i++){
                    edge&e=edges[g[x][i]];
                    if(!a[e.to]&&e.cap>e.flow){
                        p[e.to]=g[x][i];
                        a[e.to]=min(a[x],e.cap-e.flow);
                        q.push(e.to);
                    }
                }
                if(a[t]) break;
            }
            if(!a[t]) break;
            for(int u=t;u!=s;u=edges[p[u]].from){
                edges[p[u]].flow+=a[t];
                edges[p[u]^1].flow-=a[t];
            }
            flow+=a[t];
        }
        return flow;
    }
}EK;
int main()
{
    int n,m,a,b,c;
    while(scanf("%d%d",&n,&m)==2){
        EK.init(m);
        for(int i=0;i<n;i++){
            scanf("%d%d%d",&a,&b,&c);
            a--;b--;
            EK.addedge(a,b,c);
        }
        printf("%d\n",EK.Maxflow(0,m-1));
    }
    return 0;
}

//Dinic算法模板 白書358頁，點的編號從0開始
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=202;
const int inf=0x7fffffff;
struct Edge{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int>g[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void Init(int n){
        this->n=n;
        for(int i=0;i<n;i++) g[i].clear();
        edges.clear();
    }
    void Addedge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));//反向弧
        m=edges.size();
        g[from].push_back(m-2);
        g[to].push_back(m-1);
    }
    bool Bfs(){
        memset(vis,0,sizeof(vis));
        queue<int>q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty()){
            int x=q.front();q.pop();
            for(int i=0;i<(int)g[x].size();i++){
                Edge &e=edges[g[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int Dfs(int x,int a){
        if(x==t||a==0) return a;
        int flow=0,f;
        for(int&i=cur[x];i<(int)g[x].size();i++){
            Edge &e=edges[g[x][i]];
            if(d[x]+1==d[e.to]&&(f=Dfs(e.to,min(a,e.cap-e.flow)))>0){
                e.flow+=f;
                edges[g[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0) break;
            }
        }
        return flow;
    }
    int Maxflow(int s,int t){
        this->s=s;this->t=t;
        int flow=0;
        while(Bfs()){
            memset(cur,0,sizeof(cur));
            flow+=Dfs(s,inf);
        }
        return flow;
    }
}dc;
int main()
{
    int n,m,a,b,c;
    while(scanf("%d%d",&n,&m)==2){
        dc.Init(m);
        while(n--){
            scanf("%d%d%d",&a,&b,&c);
            a--;b--;
            dc.Addedge(a,b,c);
        }
        printf("%d\n",dc.Maxflow(0,m-1));
    }
    return 0;
}

//anather
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int INF=0x7fffffff;
const int MAXN=10002;//點數
const int MAXM=10002;//邊數
int n,m,tot,S,T,head[MAXN],h[MAXN],q[MAXN],ans;
struct Edge { int to,val,next; }edge[MAXM];
void init(int last)
{
    S=0;T=last-1;//S源點，T匯點
    tot=0;
    memset(head,-1,sizeof(head));
}
void addedge(int x,int y,int z)
{
    edge[tot].to=y;edge[tot].val=z;edge[tot].next=head[x];
    head[x]=tot++;
}
bool bfs()
{
    memset(h,-1,sizeof(h));
    int top=0,last=1;
    q[top]=S;h[S]=0;
    while(top<last){
        int now=q[top];top++;
        for(int i=head[now];i!=-1;i=edge[i].next){
            if(edge[i].val&&h[edge[i].to]<0){
                q[last++]=edge[i].to;
                h[edge[i].to]=h[now]+1;
            }
        }
    }
    if(h[T]==-1) return 0;
    return 1;
}
int dfs(int x,int f)
{
    if(x==T) return f;
    int w,used=0;
    for(int i=head[x];i!=-1;i=edge[i].next){
        if(edge[i].val&&h[edge[i].to]==h[x]+1){
            w=f-used;
            w=dfs(edge[i].to,min(w,edge[i].val));
            edge[i].val-=w;
            edge[i^1].val+=w;
            used+=w;
            if(used==f) return f;
        }
    }
    if(!used) h[x]=-1;
    return used;
}
int dinic()
{
    int ans=0;
    while(bfs()) ans+=dfs(S,INF);
    return ans;
}
int main()
{
    int n,m,a,b,c;
    while(scanf("%d%d",&n,&m)==2){
        init(m);//根據題目傳參
        while(n--){
            scanf("%d%d%d",&a,&b,&c);
            a--;b--;
            addedge(a,b,c);
            addedge(b,a,0);//建反向邊
            //建邊根據題目而定
        }
        int ans=dinic();
        printf("%d\n",ans);
    }
    return 0;
}