hdu 1676（二维Dijkstra）

Full Tank?

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 142    Accepted Submission(s): 64

Problem Description

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.

 

Input

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.

 

Output

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.

 

Sample Input

5 510 10 20 12 130 1 90 2 81 2 11 3 112 3 7210 0 320 1 4

 

Sample Output

170impossible

 

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

 

Recommend

lcy

 

n个城市，m条路，一个车有个有容量上限的油箱，各个城市加油的费用不同，现在求起点到终点的最小费用。因为各个城市加油的费用不同，所以简单的走最短路不一定是最优的。我们需要重新设计状态，可以想到到每个城市时的油量是有重要影响的，所以把状态设计成dp[i][j]表示到第i个城市剩余j油量的最小花费，然后状态转移是每次要么油量加1（加k的油量可以由k-1的油量的状态到达），要么走向相邻城市。注意这里的vis标记是出队的时候标记，而不是入队。

#include <iostream>#include <cstring>#include <cstdio>#include <string>#include <algorithm>#include <map>#include <vector>#include <queue>#include <set>#include <cmath>using namespace std;typedef long long LL;typedef pair<int,pair<int,int> > P;typedef pair<int,int> PP;const int maxn = 1000 + 5;const int INF = 1000000000;const int maxc = 100 + 5;int p[maxn];struct Edge{    int to,dis;    Edge(int to,int dis){        this -> to = to;        this -> dis = dis;    }};vector<Edge> G[maxn];priority_queue<P> Q;int vis[maxn][maxc];int s,e,c;int bfs(){    while(!Q.empty()){        P p1 = Q.top();Q.pop();        PP p2 = p1.second;        int price = -p1.first;        int pos = p2.first;        int gas = p2.second;        vis[pos][gas] = 1;        if(pos == e) return price;        if(gas < c && vis[pos][gas+1] == 0){            Q.push(P(-(price+p[pos]),PP(pos,gas+1)));        }        for(int i = 0;i < G[pos].size();i++){            int to = G[pos][i].to;            int dis = G[pos][i].dis;            if(dis <= gas && vis[to][gas-dis] == 0){                Q.push(P(-price,PP(to,gas-dis)));            }        }    }    return -1;}int main(){    int n,m;    while(scanf("%d%d",&n,&m) != EOF){        for(int i = 0;i < n;i++) scanf("%d",&p[i]);        for(int i = 0;i < n;i++) G[i].clear();        for(int i = 0;i < m;i++){            int u,v,d;            scanf("%d%d%d",&u,&v,&d);            G[u].push_back(Edge(v,d));            G[v].push_back(Edge(u,d));        }        int q;        scanf("%d",&q);        while(q--){            scanf("%d%d%d",&c,&s,&e);            while(!Q.empty()) Q.pop();            memset(vis,0,sizeof(vis));            Q.push(P(0,PP(s,0)));            int ans = bfs();            if(ans == -1) printf("impossible\n");            else printf("%d\n",ans);        }    }    return 0;}