UVA11367 Full Tank? 【分层图最短路】

题目

After going through the receipts from your car trip through Europe this summer,

you realised that the gas prices varied between the cities you visited. Maybe you

could have saved some money if you were a bit more clever about where you filled

your fuel?

To help other tourists (and save money yourself next time), you want to write

a program for finding the cheapest way to travel between cities, filling your tank

on the way. We assume that all cars use one unit of fuel per unit of distance, and

start with an empty gas tank.

输入格式

The first line of input gives 1 ≤ n ≤ 1000 and 0 ≤ m ≤ 10000, the number of cities and roads. Then

follows a line with n integers 1 ≤ pi ≤ 100, where pi

is the fuel price in the ith city. Then follow m lines

with three integers 0 ≤ u, v < n and 1 ≤ d ≤ 100, telling that there is a road between u and v with

length d. Then comes a line with the number 1 ≤ q ≤ 100, giving the number of queries, and q lines

with three integers 1 ≤ c ≤ 100, s and e, where c is the fuel capacity of the vehicle, s is the starting

city, and e is the goal.

输出格式

For each query, output the price of the cheapest trip from s to e using a car with the given capacity,

or ‘impossible’ if there is no way of getting from s to e with the given car.

输入样例

5 5

10 10 20 12 13

0 1 9

0 2 8

1 2 1

1 3 11

2 3 7

2

10 0 3

20 1 4

输出样例

170

impossible

题解

我们把点和在该点拥有的油量看做一个状态，建立分层图

每个点向其相邻的油量减少量为路程长的点连边【费用0】，向其自身油量+1的点连边【费用为油价】

跑dijkstra【可能SPFA跑分层图什么的会快一些】

复杂度\(O(q * n * c * log(n + c))\)

注意每一只加1升油，不能随意加油，否则就会退化到\(O(q*n*c^2*log(n + c))\)

#include<iostream>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 1005,maxm = 20005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57) {out = (out << 3) + (out << 1) + c - '0'; c = getchar();}
	return out * flag;
}
int n,m,h[maxn],ne = 2;
struct EDGE{int to,nxt,w;}ed[maxm];
void build(int u,int v,int w){
	ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;
	ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;
}
int d[maxn][105],p[maxn],vis[maxn][105];
struct node{int u,c,f;};
inline bool operator <(const node& a,const node& b){return a.f > b.f;}
priority_queue<node> q;
int solve(int S,int T,int C){
	for (int i = 1; i <= n; i++)
		for (int j = 0; j <= C; j++)
			d[i][j] = INF,vis[i][j] = 0;
	d[S][0] = 0;
	q.push((node){S,0,d[S][0]});
	int v; node u;
	while (!q.empty()){
		u = q.top(); q.pop();
		if (vis[u.u][u.c]) continue;
		vis[u.u][u.c] = true;
		Redge(u.u)
			if (ed[k].w <= u.c && !vis[to = ed[k].to][v = u.c - ed[k].w] && d[to][v] > d[u.u][u.c]){
				d[to][v] = d[u.u][u.c]; q.push((node){to,v,d[to][v]});
			}
		if (!vis[u.u][u.c + 1] && d[u.u][u.c + 1] > d[u.u][u.c] + p[u.u]){
			d[u.u][u.c + 1] = d[u.u][u.c] + p[u.u];
			q.push((node){u.u,u.c + 1,d[u.u][u.c + 1]});
		}
	}
	int ans = INF;
	for (int i = 0; i <= C; i++) ans = min(ans,d[T][i]);
	return ans;
}
int main(){
	n = read(); m = read();
	REP(i,n) p[i] = read();
	int a,b,w;
	while (m--){
		a = read() + 1; b = read() + 1; w = read();
		build(a,b,w);
	}
	int q = read(),S,T,C,ans;
	while (q--){
		C = read(); S = read() + 1; T  = read() + 1;
		ans = solve(S,T,C);
		if (ans == INF) puts("impossible");
		else printf("%d\n",ans);
	}
	return 0;
}