HDU - 6582 Path(最短路+最大流)

題目鏈接：點擊查看

題目大意：給出一張 n 個點和 m 條邊組成的有向圖，現在問讓最短路變長的最小花費是多少

題目分析：增加最短路的最小花費，我們可以將最短路上的邊單獨拿出來，再求一下最小割就好了，用了封裝後的算法看起來非常舒服，但時間複雜度有點高。。看別人都是100ms左右跑完，我的代碼是950ms劃過

代碼：
 

#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
using namespace std;
 
typedef long long LL;
 
typedef unsigned long long ull;
 
const LL inf=0x3f3f3f3f3f3f3f3f;
 
const int N=1e4+100;

template<typename T>
struct Dinic
{
	const static int N=1e4+100;
	const static int M=2e4+100;
	const T inf=0x3f3f3f3f3f3f3f3f;
	struct Edge
	{
	    int to,next;
	    T w;
	}edge[M];//邊數
	  
	int head[N],cnt;
	  
	void addedge(int u,int v,T w)
	{
	    edge[cnt].to=v;
	    edge[cnt].w=w;
	    edge[cnt].next=head[u];
	    head[u]=cnt++;
	    edge[cnt].to=u;
	    edge[cnt].w=0;//反向邊邊權設置爲0
	    edge[cnt].next=head[v];
	    head[v]=cnt++;
	}
	  
	int d[N],now[N];//深度 當前弧優化
	  
	bool bfs(int s,int t)//尋找增廣路
	{
	    memset(d,0,sizeof(d));
	    queue<int>q;
	    q.push(s);
	    now[s]=head[s];
	    d[s]=1;
	    while(!q.empty())
	    {
	        int u=q.front();
	        q.pop();
	        for(int i=head[u];i!=-1;i=edge[i].next)
	        {
	            int v=edge[i].to;
	            T w=edge[i].w;
	            if(d[v])
	                continue;
	            if(!w)
	                continue;
	            d[v]=d[u]+1;
	            now[v]=head[v];
	            q.push(v);
	            if(v==t)
	                return true;
	        }
	    }
	    return false;
	}
	  
	T dinic(int x,int t,T flow)//更新答案
	{
	    if(x==t)
	        return flow;
	    T rest=flow,i;
	    for(i=now[x];i!=-1&&rest;i=edge[i].next)
	    {
	        int v=edge[i].to;
	        T w=edge[i].w;
	        if(w&&d[v]==d[x]+1)
	        {
	            T k=dinic(v,t,min(rest,w));
	            if(!k)
	                d[v]=0;
	            edge[i].w-=k;
	            edge[i^1].w+=k;
	            rest-=k;
	        }
	    }
	    now[x]=i;
	    return flow-rest;
	}
	  
	void init()
	{
	    memset(now,0,sizeof(now));
	    memset(head,-1,sizeof(head));
	    cnt=0;
	}
	  
	T solve(int st,int ed)
	{
	    T ans=0,flow;
	    while(bfs(st,ed))
	        while(flow=dinic(st,ed,inf))
	            ans+=flow;
	    return ans;
	}
};

template<typename T>
struct Dij
{
	const static int N=1e4+100;
	const static int M=1e4+100;
	struct Edge
	{
		int to,next;
		T w;
	}edge[M];
	 
	int head[N],cnt;//鏈式前向星 
	
	T d[N];
	 
	bool vis[N];
	 
	void addedge(int u,int v,T w)
	{
		edge[cnt].to=v;
		edge[cnt].w=w;
		edge[cnt].next=head[u];
		head[u]=cnt++;
	}
	 
	struct Node
	{
		int to;
		T w;
		Node(int TO,T W)
		{
			to=TO;
			w=W;
		}
		bool operator<(const Node& a)const
		{
			return w>a.w;
		}
	};
	 
	void Dijkstra(int st)
	{
		priority_queue<Node>q;
		memset(vis,false,sizeof(vis));
		memset(d,0x3f,sizeof(d));
		d[st]=0;
		q.push(Node(st,0));
		while(q.size())
		{
			Node cur=q.top();
			int u=cur.to;
			q.pop();
			if(vis[u])
				continue;
			vis[u]=true;
			for(int i=head[u];i!=-1;i=edge[i].next)//掃描出所有邊 
			{
				int v=edge[i].to;
				T w=edge[i].w;
				if(d[v]>d[u]+w)//更新 
				{
					d[v]=d[u]+w;
					q.push(Node(v,d[v]));
				}
			}
		}
	}
	 
	void init()
	{
		memset(head,-1,sizeof(head));
		cnt=0; 
	}
};

Dinic<LL>t;

Dij<LL>d1,d2;

int main()
{
#ifndef ONLINE_JUDGE
//  freopen("input.txt","r",stdin);
//  freopen("output.txt","w",stdout);
#endif
//  ios::sync_with_stdio(false);
	int w;
	cin>>w;
	while(w--)
	{
		t.init(),d1.init(),d2.init();
		int n,m,st=N-1,ed=st-1;
		scanf("%d%d",&n,&m);
		while(m--)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			d1.addedge(u,v,w);
			d2.addedge(v,u,w);
		}
		d1.Dijkstra(1);
		d2.Dijkstra(n);
		if(d1.d[n]==inf)
		{
			puts("0");
			continue;
		}
		for(int i=1;i<=n;i++)
			for(int j=d1.head[i];j!=-1;j=d1.edge[j].next)
			{
				int u=i,v=d1.edge[j].to;
				LL w=d1.edge[j].w;
				if(d1.d[u]+w+d2.d[v]==d1.d[n])
					t.addedge(u,v,w);
			}
		t.addedge(st,1,inf);
		t.addedge(n,ed,inf);
		printf("%lld\n",t.solve(st,ed));
	}





















    return 0;
}