HDU 6290 奢侈的旅行

題目

題目的思路很簡單，又式子可得就是求到達n的時候level最小是多少，又由通過路徑的要求可以得，level越大越不可能通過這條路徑，所以就是一個跑一遍單源最短路就可以了。

默唸三遍

priority_queue<int> 是大根堆

priority_queue<int, vector<int>, greater<int> > 是小根堆 

以及精度問題

用（log(w+a)-log(w))/log(2)瘋狂WA題，用log((w+a)/w)/log(2)就AC了55555555555

#include<iostream>
#include<stdio.h> 
#include<algorithm>
#include<string.h>
#include<vector>
#include<set>
#include<math.h>
#include<queue>
#include<map>
#include<stack>
#define go(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define ll long long
#define N 100005
#define INF 
using namespace std;
typedef pair<ll,ll> pll;
struct edge{
	ll v,a,b,next;
}e[N*2];
ll h[N],tot;
void add(ll u, ll v, ll a, ll b){
	e[++tot]=(edge){v,a,b,h[u]};
	h[u]=tot;
}
ll dis[N];
bool vis[N];
priority_queue<pll, vector<pll>, greater<pll> >q;

void dij(ll n, ll st){
	memset(dis,-1,sizeof(dis));
	memset(vis,false,sizeof(vis));
	dis[st]=1;
	while (!q.empty()) q.pop();
	q.push(make_pair(1,st));
	while (!q.empty()){
		pll tmp=q.top();
		q.pop();
		ll w=tmp.first;
		ll u=tmp.second;
		if (vis[u]) continue;
		vis[u]=true;
		for (ll i=h[u];i!=-1;i=e[i].next){
			ll v=e[i].v;
			if (!vis[v]&&(log((w+e[i].a)/w))/log(2)>=e[i].b&&(dis[v]>w+e[i].a||dis[v]==-1)){
				dis[v]=w+e[i].a;
				q.push(make_pair(w+e[i].a,v));
			}
		}
	}
}
int main(){
	ll T,n,m,u,v,a,b;
	scanf("%lld",&T);
	while (T--){
		scanf("%lld%lld",&n,&m);
		memset(h,-1,sizeof(h));tot=0;
		go(i,1,m){
			scanf("%lld%lld%lld%lld",&u,&v,&a,&b);
			add(u,v,a,b);
		}
		dij(n,1);
		//go(i,1,n) cout<<i<<" "<<dis[i]<<endl;
		if (dis[n]==-1) printf("-1\n");
		else printf("%lld\n",(ll)(log(dis[n])/log(2)));
	}
}


/*
1
3 3
1 2 3 2
2 3 4 0
1 3 5 6


*/