poj3635Full Tank? 广搜最短路
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5754 | Accepted: 1891 |
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 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
Sample Output
170 impossible--------------
【题意】
给你 n 个点，m 条边，每走 1 单位的路径都会花费 1 单位的 fuel ，并且不同的点灌油的油的价格是不同的，现在给你一些询问，每一个询问给你起点、终点以及油箱的容量，问你所需要的最少的花费可以从起点到达终点。
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/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) { \
REP(i, n){ \
REP(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#define Display_1(A, n, m) { \
REP_1(i, n){ \
REP_1(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=2111;
const int maxm=31111;
int n,m;
struct EDGENODE{
int to;
int w;
int next;
};
struct SGRAPH{
int head[maxn];
EDGENODE edges[maxm];
int edge;
void init(){
clr(head,-1);
edge=0;
}
void addedge(int u,int v,int c=0){
edges[edge].w=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
}dragonborn;
struct data{
int u,f,c;
data(){}
data(int u,int f,int c){
this->u=u;
this->f=f;
this->c=c;
}
friend bool operator<(const data& a,const data& b)
{
return a.c>b.c;
}
};
int cost[maxn];
bool vis[maxn][111];
int solve(SGRAPH& G,int V,int s,int e)
{
priority_queue<data>que;
clr(vis,0);
que.push(data(s,0,0));
while (!que.empty())
{
data top=que.top();
que.pop();
vis[top.u][top.f]=true;
if (top.u==e) return top.c;
if (top.f<V&&!vis[top.u][top.f+1])
que.push(data(top.u,top.f+1,top.c+cost[top.u]));
for (int i=G.head[top.u];i!=-1;i=G.edges[i].next)
{
if (top.f>=G.edges[i].w&&!vis[G.edges[i].to][top.f-G.edges[i].w])
que.push(data(G.edges[i].to,top.f-G.edges[i].w,top.c));
}
}
return -1;
}
int main()
{
while (~scanf("%d%d",&n,&m))
{
int Q;
dragonborn.init();
REP(i,n) scanf("%d",&cost[i]);
REP(i,m)
{
int u,v,d;
scanf("%d%d%d",&u,&v,&d);
dragonborn.addedge(u,v,d);
dragonborn.addedge(v,u,d);
}
scanf("%d",&Q);
while (Q--)
{
int c,s,e;
scanf("%d%d%d",&c,&s,&e);
int ret=solve(dragonborn,c,s,e);
if (ret!=-1) printf("%d\n",ret);
else printf("impossible\n");
}
}
return 0;
}
