网络流dinic ek模板 poj1273

Posted 2020-10-09

这里只是用来存放模板，几乎没有讲解，要看讲解网上应该很多吧……

ek

bfs不停寻找增广路到找不到为止，找到终点时用pre回溯，O(VE^2)

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 205;
int cap[205][205], pre[205], n, m, a[205];
int bfs(){
    queue<int> q;
    q.push(1);
    memset(a,0,sizeof(a));
    a[1] = INF;
    while(!q.empty()){
        int front = q.front();
        q.pop();
        for(int i = 1;i<=m;i++){
            if(!a[i] && cap[front][i]){
                a[i] = min(a[front], cap[front][i]);
                pre[i] = front;
                q.push(i);
            }
        }
    }
    return a[m];
}
int ek(){
    int ans = 0;
    while(bfs()){
        ans += a[m];
        for(int i = m;i!=1;i = pre[i]){
            cap[pre[i]][i] -= a[m];
            cap[i][pre[i]] += a[m];
        }
    }
    return ans;
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        memset(cap,0,sizeof(cap));
        for(int i = 1;i<=n;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            cap[u][v] += w;
        }
        printf("%d\n",ek());    
    }
    return 0;    
}

dinic

反复构造层次网络加找增广路，优势在于当某点的流入量较大时，可以一次完成多条增广路的累加，O(V^2E)

记得初始化lv，cnt=1

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 205;
const int INF = 0x3f3f3f3f;
struct EDGE{
    int to, next, cap, flow;
    EDGE(){}
    EDGE(int a, int b, int c, int d){
        to = a, next = b, cap = c, flow = d;;
    }
}e[410];
int head[205], cnt, lv[205],m ,n;
void add(int from, int to, int cap){
    e[++cnt] = EDGE(to, head[from], cap, 0);
    head[from] = cnt;
    e[++cnt] = EDGE(from, head[to], 0, 0);
    head[to] = cnt;
}
int bfs(){
    queue<int> q;
    q.push(1);
    memset(lv,0,sizeof(lv));
    lv[1] = 1;
    while(!q.empty()){
        int front = q.front();
        q.pop();
        for(int i = head[front];i;i = e[i].next){
            int to = e[i].to;
            if(!lv[to] && e[i].cap-e[i].flow){
                lv[to] = lv[front]+1;
                q.push(to);
            }
        }
    }
    return lv[m];
}
int dfs(int now, int a){
    int flow = 0,f;
    if(now == m) return a;
    for(int i = head[now];i;i = e[i].next){
        int to = e[i].to;
        if(lv[to] == lv[now]+1 && (f = dfs(to,min(a,e[i].cap-e[i].flow)))){
            e[i].flow += f;
            e[i^1].flow -= f;
            flow += f;
            a -= f;
            if(!a) break;
        }
    }
    return flow;
} 
int dinic(){
    int ans = 0;
    while(bfs()){
        ans += dfs(1,INF);
    }
    return ans;
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        cnt = 1;
        memset(head,0,sizeof(head));
        for(int i = 1;i<=n;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
        }
        printf("%d\n",dinic());    
    }
    return 0;    
}