学步园

题目链接：http://acm.nbu.edu.cn/v1.0/Problems/Problem.php?pid=2430

题目大意：

有n个城市（编号1~n），m条管道，每条管道连接两个城市，管道i最多可容纳c[i]的能量通过，求从1最多能向n传送多少能量。

0<=n<=200,0<=m<=40000,0<=ci<=100000.

题目思路：

裸的最大流，按照题意建图，1为源点，n为汇点，求最大流。

代码：

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll __int64
#define ls rt<<1
#define rs ls|1
#define lson l,mid,ls
#define rson mid+1,r,rs
#define middle (l+r)>>1
#define eps (1e-8)
#define clr_all(x,c) memset(x,c,sizeof(x))
#define clr(x,c,n) memset(x,c,sizeof(x[0])*(n+1))
#define MOD (1000000007)
#define inf (0x3f3f3f3f)
#define pi (acos(-1.0))
#define _max(x,y) (((x)>(y))? (x):(y))
#define _min(x,y) (((x)<(y))? (x):(y))
#define _abs(x) ((x)<0? (-(x)):(x))
#define getmin(x,y) ((x)= ((x)<0 || (y)<(x))? (y):(x))
#define getmax(x,y) ((x)= ((y)>(x))? (y):(x))
template <class T> void _swap(T &x,T &y){T t=x;x=y;y=t;}
int TS,cas=1;
const int M=200+5;
int n,m;
struct sap{
	typedef ll type;
	struct edge{
		int to,next;
		type flow;
	}edg[999999];
	int n,m;
	int g[M],cur[M],pre[M];
	int dis[M],gap[M];
	int q[M],head,tail;
	void init(int _n){n=_n,m=0,clr_all(g,-1);}
	void insert(int u,int v,type f,type c=0){
		edg[m].to=v,edg[m].flow=f,edg[m].next=g[u],g[u]=m++;
		edg[m].to=u,edg[m].flow=0,edg[m].next=g[v],g[v]=m++;
	}
	bool bfs(int s,int t){
		clr(gap,0,n),clr(dis,-1,n);
		gap[dis[t]=0]++,dis[s]=-1;
		for(q[head=tail=0]=t;head<=tail;){
			int u=q[head++];
			for(int i=g[u];i!=-1;i=edg[i].next){
				edge& e=edg[i],ee=edg[i^1];
				if(dis[e.to]==-1 && ee.flow>0){
					gap[dis[e.to]=dis[u]+1]++;
					q[++tail]=e.to;
				}
			}
		}
		return dis[s]!=-1;
	}
	type maxFlow(int s,int t){
		if(!bfs(s,t)) return 0;
		type res=0,a;
		int i;
		for(i=0;i<n;i++) cur[i]=g[i];
		pre[s]=s,cur[s]=g[s],cur[t]=g[t];
		for(int u=s;dis[s]<n;){
			if(u==t){
				for(a=-1;u!=s;u=pre[u])
					getmin(a,edg[cur[pre[u]]].flow);
				for(u=t;u!=s;u=pre[u]){
					edg[cur[pre[u]]].flow-=a;
					edg[cur[pre[u]]^1].flow+=a;
				}
				res+=a;
			}
			bool ok=0;
			for(i=cur[u];i!=-1;i=edg[i].next){
				edge& e=edg[i];
				if(dis[u]==dis[e.to]+1 && e.flow>0){
					cur[u]=i,pre[e.to]=u,u=e.to;
					ok=1;break;
				}
			}
			if(ok) continue;
			int mindis=n-1;
			for(i=g[u];i!=-1;i=edg[i].next){
				edge& e=edg[i];
				if(mindis>dis[e.to] && e.flow>0)
					mindis=dis[e.to],cur[u]=i;
			}
			if(--gap[dis[u]]==0) break;
			gap[dis[u]=mindis+1]++,u=pre[u];
		}
		return res;
	}
}p;

void run(){
    int i,j;
	p.init(n);
	while(m--){
		int s,e;
		ll c;
		scanf("%d%d%I64d",&s,&e,&c);
		p.insert(s-1,e-1,c);
	}
	printf("%I64d\n",p.maxFlow(0,n-1));
}

void preSof(){
}

int main(){
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    preSof();
    //run();
    while(~scanf("%d%d",&n,&m)) run();
    //for(scanf("%d",&TS);cas<=TS;cas++) run();
    return 0;
}