【PA2012】【BZOJ4289】Tax

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

Input

Output

Sample Input

4 5

1 2 5

1 3 2

2 3 1

2 4 4

3 4 8

Sample Output

12

HINT

Source

一眼能看出是最短路变种..

可是我仅仅会n^2建图T_T

n^2建图就是边转点随便建个新图然后最短路即可了..

这妥妥的TLE啊_ (:зゝ∠)_

这里是Claris的正解

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define MAXN 100010
#define MAXM 200010
#define LL long long
#define GET (ch>=‘0‘&&ch<=‘9‘)
using namespace std;
int n,m,top,Top,cnt,_top;
int start,end;
const LL MAXINT=1ll<<60;
LL dis[MAXM<<1];
bool vis[MAXM<<1];
struct edge
{
    int to,st;
    edge *next;
}e[MAXM<<1],*prev[MAXN];
void insert(int u,int ed,int st)    {e[++top].to=ed;e[top].st=st;e[top].next=prev[u];prev[u]=&e[top];}
struct Edge
{
    int to,w;
    Edge *next;
}E[MAXM<<3],*Prev[MAXM<<1];
void Insert(int u,int v,int w)  {E[++Top].to=v;E[Top].w=w;E[Top].next=Prev[u];Prev[u]=&E[Top];}
struct S
{
    int u,v,w;
    bool operator <(const S& a)const    {return w<a.w;}
}s[MAXM<<1],sta[MAXM];
void in(int &x)
{
    char ch=getchar();x=0;
    while (!GET)    ch=getchar();
    while (GET) x=x*10+ch-‘0‘,ch=getchar();
}
struct node
{
    int x;
    LL dis;
    bool operator <(const node& a)const {return dis>a.dis;}
};
void dijkstra(int s)
{
    priority_queue<node>    heap;
    for (int i=1;i<=end;i++)    dis[i]=MAXINT;dis[s]=0;heap.push((node){s,0});
    while (!heap.empty())
    {
        int x=heap.top().x;heap.pop();
        if (vis[x]) continue;vis[x]=1;
        for (Edge *i=Prev[x];i;i=i->next)
            if (dis[i->to]>dis[x]+i->w) dis[i->to]=dis[x]+i->w,heap.push((node){i->to,dis[i->to]});
    }
}
int main()
{
    in(n);in(m);int u,v,w;
    for (int i=1;i<=m;i++)
    {
        in(u);in(v);in(w);
        s[++cnt]=(S){u,v,w};s[++cnt]=(S){v,u,w};
        insert(u,cnt,cnt-1);insert(v,cnt-1,cnt);
    }
    for (int i=1;i<=n;i++)
    {
        _top=0;
        for (edge *j=prev[i];j;j=j->next)   sta[++_top]=(S){j->to,j->st,s[j->to].w};
        if (!_top)  continue;sort(sta+1,sta+_top+1);
        for (int j=1;j<=_top;j++)   Insert(sta[j].u,sta[j].v,sta[j].w);
        for (int j=1;j<_top;j++)    Insert(sta[j].v,sta[j+1].v,sta[j+1].w-sta[j].w),Insert(sta[j+1].v,sta[j].v,0);
    }
    start=cnt+1;end=start+1;
    for (int i=1;i<=cnt;i++)
    {
        if (s[i].u==1)  Insert(start,i,s[i].w);
        if (s[i].v==n)  Insert(i,end,s[i].w);
    }
    dijkstra(start);cout<<dis[end]<<endl;
}