P2149 Elaxia的路线

P2149 Elaxia的路线

 

题意简述： 在一个n(n<=1500)个点的无向图里找两对点之间的最短路径的最长重合部分，即在保证最短路的情况下两条路径的最长重合长度(最短路不为一)

 

• 思路：
• 两边dij，第一遍最短路，第二遍在保证最短路的情况下让重合路径长度最长的先出队
• 代码：

#include <cstdio>
#include <iostream>
#include <queue>
#include <cctype>
#include <cstring>
#include <algorithm>
using namespace std;

#define res register int 
inline int read()
{
	int x=0,f=1; char ch;
	while(!isdigit(ch=getchar()))if(ch=='-')f=-1;
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f*x;
}

const int N=1500+10,M=N*N*2;
int x1,y1,x2,y2,n,m,tot=1;

struct node{
	int x,y;//xΪ×î¶Ì·³¤¶È£¬yΪ×ÖغÏ·¾¶³¤¶È 
	bool operator >(const node &n2) const{return x==n2.x?y<n2.y:x>n2.x;}
	bool operator <(const node &n2) const{return x==n2.x?y>n2.y:x<n2.x;}//СÓÚ ->¸üÓÅ
	node(int _x=0,int _y=0) : x(_x),y(_y){}
	bool operator ==(const node &n2) const{return x==n2.x&&y==n2.y;}
	node operator +(const node &n2) const{return node(x+n2.x,y+n2.y);} 
}edge[M],dis[M];
int ver[M],nxt[M],head[N],vis[N];

inline void add(int x,int y,int z)
{
	ver[++tot]=y; nxt[tot]=head[x]; head[x]=tot; edge[tot]=node(z,0);
}
	
struct point{
	int p; node d;
	bool operator <(const point &n2) const{return d>n2.d;}
	point(int _p=0,node _d=node(0,0)) : p(_p),d(_d) {}
};

inline void dij(int s)
{
	priority_queue <point> q;
	for(res i=1 ; i<=n ; ++i) dis[i]=node(0x3f3f3f3f,0),vis[i]=0;
	dis[s]=node(0,0); q.push(point(s,dis[s]));
	while(q.size())
	{
		point now=q.top(); q.pop();
		int x=now.p;
		if(vis[x]) continue; vis[x]=1;
		for(res i=head[x] ; i ; i=nxt[i])
		{
			int y=ver[i];
			if(dis[y]>dis[x]+edge[i])
			{
				dis[y]=dis[x]+edge[i];
				q.push(point(y,dis[y]));
			}
		}
	}
}

void dfs(int x)
{
	for(res i=head[x] ; i ; i=nxt[i])
	{
		int y=ver[i];
		if(dis[y]+edge[i]==dis[x])
		{
			edge[i].y+=edge[i].x; edge[i^1].y+=edge[i].x;
			dfs(y);
		}
	}
}

int main()
{
	n=read(); m=read();
	x1=read(); y1=read(); x2=read(); y2=read();
	for(res i=1 ; i<=m ; ++i)
	{
		int x=read(),y=read(),z=read(); add(x,y,z); add(y,x,z);
	}
	dij(x1); dfs(y1); dij(x2); 
	printf("%d\n",dis[y2].y);
	return 0;
}