hdu 3549 Flow Problem Edmonds_Karp算法求解最大流

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..

Edmond_Karp算法BFS查找每次需要O(m)总时间复杂度为O(n*m2),不够快所以跑了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=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';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>) out(a/);
putchar(a%+'');
}
const int M = ;
const int N = ;
int head[M<<],tot;
struct edge{
int from,to,cap,flow,Next;
}e[M<<];
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 = ;
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]^].flow -= a[t];
}
flow += a[t];
}
return flow;
}
int main()
{
int n,T,kase = ;
read1(T);
while(T--){
int V,E;
read2(V,E);
MS1(head);tot = ;
rep0(i,,E){
int u,v,w;
read3(u,v,w);
ins(u,v,w,);ins(v,u,,);
}
printf("Case %d: ",kase++);
out(Edmonds_Karp(,V));
puts("");
}
return ;
}
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