BZOJ4289 PA2012 Tax

Problem

链接

Description

给出一个N个点M条边的无向图，经过一个点的代价是进入和离开这个点的两条边的边权的较大值，求从起点1到点N的最小代价。起点的代价是离开起点的边的边权，终点的代价是进入终点的边的边权

N<=100000

M<=200000

Sample Input

4 5
1 2 5
1 3 2
2 3 1
2 4 4
3 4 8

Sample Output

12

Solution

//lzjtxdy
#include<iostream>
#include<cstdio> 
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int N=2000000,M=2000000;
int head[N],ver[M],nxt[M],edge[M],tot=0;
int Head[N],Ver[M],Nxt[M],Edge[M],Tot=0;
bool book[N];
long long dis[N];
#define s(x) ((x)%2==1?((x)+1):((x)-1))
struct node
{
    int z;
    int pos;
}tmp[N];
void add(int x,int y,int z)
{
    ver[++tot]=y;
    edge[tot]=z;
    nxt[tot]=head[x];
    head[x]=tot;
}
void addedge(int x,int y,int z)
{
	Ver[++Tot]=y;
	Edge[Tot]=z;
	Nxt[Tot]=Head[x];
	Head[x]=Tot;
}
void dijkstra(int st)
{
	priority_queue<pair<long long,int> > que;
	que.push(make_pair(0,st));
	memset(dis,0x3f,sizeof(dis));
	dis[st]=0;
	memset(book,false,sizeof(book));
	while(!que.empty())
	{
		int x=que.top().second; que.pop();
		if(book[x]) continue;
		book[x]=true;
		for(int i=head[x];i;i=nxt[i])
		{
			int y=ver[i],z=edge[i];
			if(dis[x]+z<dis[y])
			{
				dis[y]=dis[x]+z;
				que.push(make_pair(-dis[y],y));
			}
		}
	}
}
bool cmp(node a,node b) {return a.z<b.z;}
void Add(int x)
{
    int cnt=0;
    for(int i=Head[x];i;i=Nxt[i])
    {
    	tmp[++cnt].pos=i;
    	tmp[cnt].z=Edge[i];
	}
	sort(tmp+1,tmp+cnt+1,cmp);
	for(int i=1;i<cnt;i++)
	{
		add(tmp[i].pos,tmp[i+1].pos,tmp[i+1].z-tmp[i].z);
		add(tmp[i+1].pos,tmp[i].pos,0);
	}
	for(int i=Head[x];i;i=Nxt[i]) add(s(i),i,Edge[i]);
}
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=m;i++)
    {
    	int xx,yy,zz;
    	scanf("%d %d %d",&xx,&yy,&zz);
    	addedge(xx,yy,zz);
    	addedge(yy,xx,zz);
	}
	int S=0,T=Tot+1;
	for(int i=Head[1];i;i=Nxt[i]) add(S,i,Edge[i]);
	for(int i=Head[n];i;i=Nxt[i]) add(s(i),T,Edge[i]);
	for(int i=1;i<=n;i++) Add(i);
	dijkstra(S);
	printf("%lld",dis[T]);
	return 0;
}

标签：int,Tax,PA2012,tot,que,Tot,include,BZOJ4289,dis
来源： https://www.cnblogs.com/juruo-zzt/p/12810924.html