POJ1273Drainage Ditches

链接：http://poj.org/problem?id=1273

题意：给定n条边和m个点的图，有边权。求从1到m的最大流量。

分析：学习网络流的第一题，dinic算法的裸题。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=210;
const int MAX=1000000100;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const int INF=1000000010;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned long long ull;
int tot,u
,v[2*N],flow[2*N],pre[2*N];
void add(int a,int b,int f) {
v[tot]=b;flow[tot]=f;pre[tot]=u[a];u[a]=tot++;
}
bool q
;
int d
,dis
,head[2*N];
bool bfs(int n) {
int i,l=1,r=1;
memset(q,0,sizeof(q));
q[1]=1;d[1]=1;dis[1]=0;
for (;l<=r;l++) {
for (i=u[d[l]];i!=-1;i=pre[i])
if (!q[v[i]]&&flow[i]) d[++r]=v[i],q[v[i]]=1,dis[v[i]]=dis[d[l]]+1;
}
return q
;
}
int dfs(int x,int a,int en) {
if (x==en||a==0) return a;
int ret=0,f;
for (int& i=u[x];i!=-1;i=pre[i])
if (dis[v[i]]==dis[x]+1&&(f=dfs(v[i],min(a,flow[i]),en))>0) {
flow[i]-=f;flow[i^1]+=f;
ret+=f;a-=f;
if (a==0) break ;
}
return ret;
}
int main()
{
int a,b,f,i,n,m,ans,xx=0;
while (scanf("%d%d", &m, &n)!=EOF) {
tot=0;ans=0;
memset(u,-1,sizeof(u));
for (i=1;i<=m;i++) {
scanf("%d%d%d", &a, &b, &f);
add(a,b,f);add(b,a,0);
}
for (i=1;i<=n;i++) head[i]=u[i];
while (bfs(n)) {
ans+=dfs(1,INF,n);
for (i=1;i<=n;i++) u[i]=head[i];
}
printf("%d\n", ans);
}
return 0;
}