hdu 3549 Flow Problem

Posted 2020-06-13

Flow Problem

题意:N个顶点M条边，(2 <= N <= 15, 0 <= M <= 1000)问从1到N的最大流量为多少？

分析:直接使用Edmonds_Karp算法即可；下面是对增广路的一些理解和代码的解释；

残量:容量-流量；

增广：求出从源点到汇点的一条道路中所有残量的最小值d,把对应的所有边上的流量增加d,反向边(t->s)流量减少d(反向边的cap其实一直是0，只是flow为负了);

技巧:这次的ins的标号是从0开始的，即tot++,之前我都是++tot;这样head初始化就变为-1了，不能再是0；这样是为了每条边和其反向边的编号是存在XOR关系；即每次找到一条道路后从t找回到s(所以要在边中加入from)对每条边及其反向边的残量变化；

注意:同时增广路可达到所有边数的两倍；以及每次寻找路径的时候要把queue清空，否则MLE..

// 218ms
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<set>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define inf 0x3f3f3f3f
#define pb push_back
template<typename T>
void read1(T &m)
{
    T x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+‘0‘);
}
const int M = 1000;
const int N = 20;
int head[M<<1],tot;
struct edge{
    int from,to,cap,flow,Next;
}e[M<<1];
void ins(int u,int v,int cap,int flow)
{
    e[tot].Next = head[u];
    e[tot].from = u;//为了t->s时由v推到u;
    e[tot].to = v;
    e[tot].cap = cap;
    e[tot].flow = flow;
    head[u] = tot++;
}
queue<int> q;
int p[N];//记录路径中边的标号
int a[N];//起点到i的可改进量
int Edmonds_Karp(int s,int t)
{
    int flow = 0;
    for(;;){
        MS0(a);
        while(!q.empty()) q.pop();
        a[s] = inf;
        q.push(s);
        while(!q.empty()){
            int u = q.front();q.pop();
            for(int id = head[u];~id;id = e[id].Next){
                int v = e[id].to,c = e[id].cap,f = e[id].flow;
                if(!a[v] && c > f){
                    p[v] = id;
                    a[v] = min(a[u],c - f);// ** 递推到a[v]
                    q.push(v);
                }
            }
            if(a[t]) break;
        }
        if(!a[t]) break;
        for(int u = t;u != s;u = e[p[u]].from){
            e[p[u]].flow += a[t];
            e[p[u]^1].flow -= a[t];
        }
        flow += a[t];
    }
    return flow;
}
int main()
{
    int n,T,kase = 1;
    read1(T);
    while(T--){
        int V,E;
        read2(V,E);
        MS1(head);tot = 0;
        rep0(i,0,E){
            int u,v,w;
            read3(u,v,w);
            ins(u,v,w,0);ins(v,u,0,0);
        }
        printf("Case %d: ",kase++);
        out(Edmonds_Karp(1,V));
        puts("");
    }
    return 0;
}