POJ 3635 Full Tank? 最短路DP

给出一个图(1≤n≤1000,0≤m≤10000)，每个点有加油站费用ci，对于每个询问，油箱容积c的车从s到e的最小费用。

如果这个图是个DAG的话很显然我们可以直接使用动态规划求解。

但它现在不是DAG，回忆起GD神犇姜碧野（应该没记错。。）有篇论文是讲SPFA的应用的，不过没负边，用Dij+Heap即可求有环的动态规划。

#include <cstdio>
#include <cstring>
#include <queue>
#define rep(i,j,k) for(i=j;i<k;i++)
using namespace std;
const int N = 1001, M = 20001, inf = 0x7f7f7f7f;
int h
, p[M], v[M], w[M], c
, cnt = 0;
int dp
[101], vis
[101], s, t, cap;
struct Data { int x, o, c; };
bool operator <(Data a, Data b){ return a.c>b.c; }

void add(int x, int y, int z) {
p[++cnt] = h[x]; w[cnt] = z; v[cnt] = y; h[x] = cnt;
}

int dij() {
priority_queue<Data> q;
memset(dp, 127, sizeof dp);
memset(vis, 0, sizeof vis);
dp[s][0] = 0;
q.push((Data) { s, 0 });
while (!q.empty()) {
Data u = q.top(); q.pop();
vis[u.x][u.o] = 1;
if (u.x == t) return u.c;
if (u.o < cap && !vis[u.x][u.o + 1] &&
dp[u.x][u.o + 1] > dp[u.x][u.o] + c[u.x]) {
dp[u.x][u.o + 1] = dp[u.x][u.o] + c[u.x];
q.push((Data) {u.x, u.o + 1, dp[u.x][u.o + 1]});
}
for (int i = h[u.x]; i; i = p[i])
if (u.o >= w[i] && !vis[v[i]][u.o - w[i]] &&
dp[v[i]][u.o - w[i]] > dp[u.x][u.o]) {
dp[v[i]][u.o - w[i]] = dp[u.x][u.o];
q.push((Data) {v[i], u.o - w[i], dp[v[i]][u.o - w[i]]});
}
}
return -1;
}

int main() {
int n, i, m, q;
scanf("%d%d", &n, &m);
rep(i,0,n) scanf("%d", c + i);
while (m--) {
scanf("%d%d%d", &s, &t, &cap);
add(s, t, cap); add(t, s, cap);
}
scanf("%d", &q);
while (q--) {
scanf("%d%d%d", &cap, &s, &t);
int re = dij();
if (re == -1) puts("impossible");
else printf("%d\n", re);
}
return 0;
}

Full Tank?

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 6910 Accepted: 2248

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