Full Tank?
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7543 | Accepted: 2444 |
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
Source
【题解】
设b[i][j]表示在第i个城市，剩余汽油j的答案
每次选答案最小的去拓展，拓展方案无非两种：买1个汽油或者走向下一个城市
注意如果新的方案答案比已有答案小，就不再加入优先队列了
其实出队的节点b[i][j]，一定是最优的
很（一点也不）显然，没有更小的值去更新b[i][j]了，只能用b[i][j]去更新别人
很想dijkstra
复杂度O(n*clogn*c)
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstring>
4 #include <cstdio>
5 #include <algorithm>
6 #include <queue>
7
8 inline void read(int &x)
9 {
10 x = 0;char ch = getchar(), c = ch;
11 while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();
12 while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
13 if(c == ‘-‘)x = -x;
14 }
15
16 const int MAXN = 2000 + 10;
17 const int MAXM = 20000 + 10;
18 const int MAXC = 200 + 10;
19
20 struct Edge
21 {
22 int u,v,w,next;
23 Edge(int _u, int _v, int _w, int _next){u = _u;v = _v; w = _w; next = _next;}
24 Edge(){}
25 }edge[MAXM << 2];
26 int head[MAXN], cnt;
27
28 inline void insert(int a, int b, int c)
29 {
30 edge[++cnt] = Edge(a,b,c,head[a]);
31 head[a] = cnt;
32 }
33
34 int n,m,value[MAXN],qq,b[MAXN][MAXC],s,e,c;
35
36 struct Node
37 {
38 int now, v, ans;
39 Node(){}
40 Node(int _now, int _v, int _ans){now = _now;v = _v; ans = _ans;}
41 };
42
43 struct cmp
44 {
45 bool operator()(Node a, Node b)
46 {
47 return a.ans > b.ans;
48 }
49 };
50
51 std::priority_queue<Node, std::vector<Node>, cmp> q;
52
53 void bfs(int s, int e)
54 {
55 memset(b, 0x3f, sizeof(b));
56 while(q.size())q.pop();
57 q.push(Node(s, 0, 0));
58 b[0][0] = 1;
59 if(s == e)
60 {
61 printf("0\n");
62 return;
63 }
64 register Node tmp;
65 while(q.size())
66 {
67 tmp = q.top();q.pop();
68 if(tmp.v + 1 <= c && b[tmp.now][tmp.v + 1] > tmp.ans + value[tmp.now])q.push(Node(tmp.now, tmp.v + 1, tmp.ans + value[tmp.now])), b[tmp.now][tmp.v + 1] = tmp.ans + value[tmp.now];
69 for(register int pos = head[tmp.now];pos;pos = edge[pos].next)
70 {
71 if(tmp.v < edge[pos].w || b[edge[pos].v][tmp.v - edge[pos].w] <= tmp.ans)continue;
72 if(edge[pos].v == e)
73 {
74 printf("%d\n", tmp.ans);
75 return;
76 }
77 q.push(Node(edge[pos].v, tmp.v - edge[pos].w, tmp.ans));
78 b[edge[pos].v][tmp.v - edge[pos].w] = tmp.ans;
79 }
80 }
81 printf("impossible\n");
82 }
83
84 int main()
85 {
86 read(n);read(m);
87 register int tmp1, tmp2, tmp3;
88 for(register int i = 1;i <= n;++ i)read(value[i]);
89 for(register int i = 1;i <= m;++ i)
90 {
91 read(tmp1), read(tmp2), read(tmp3);
92 insert(tmp1 + 1, tmp2 + 1, tmp3);
93 insert(tmp2 + 1, tmp1 + 1, tmp3);
94 }
95 read(qq);
96 for(;qq;-- qq)
97 {
98 read(c);read(s);read(e);
99 ++ s;++ e;
100 bfs(s, e);
101 }
102 return 0;
103 }
POJ3635