1003. Emergency (25)
As an emergency rescue team leader of a city, you are given a special map of
your country. The map shows several scattered cities connected by some roads.
Amount of rescue teams in each city and the length of each road between any
pair of cities are marked on the map. When there is an emergency call to you
from some other city, your job is to lead your men to the place as quickly as
possible, and at the mean time, call up as many hands on the way as
possible.
Input
Each input file contains one test case. For each test case, the first line
contains 4 positive integers: N (<= 500) - the number of cities (and the
cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the
cities that you are currently in and that you must save, respectively. The
next line contains N integers, where the i-th integer is the number of rescue
teams in the i-th city. Then M lines follow, each describes a road with three
integers c1, c2 and L, which are the pair of cities connected by a road and the
length of that road, respectively. It is guaranteed that there exists at
least one path from C1 to C2.
Output
For each test case, print in one line two numbers: the number of different
shortest paths between C1 and C2, and the maximum amount of rescue teams you
can possibly gather.
All the numbers in a line must be separated by exactly
one space, and there is no extra space allowed at the end of a
line.
Sample Input5 6 0 2 1 2 1 5 3 0 1 1 0 2 2 0 3 1 1 2 1 2 4 1 3 4 1
Sample
Output2 4
1 #include <stdio.h>
2 #include <string.h>
3 #include <limits.h>
4
5 int dijkstra(int,int,int&,int&);
6 int N,M,c1,c2;
7 int num_rescue[500];
8 int map[500][500];
9 int cost[500];
10 int num_path[500];
11 int num_res_gather[500];
12 int flag[500];
13 int main()
14 {
15 int i;
16 int s,e,c;
17 while (scanf("%d%d%d%d",&N,&M,&c1,&c2) != EOF){
18 memset(num_rescue,0,sizeof(num_rescue));
19 memset(map,0,sizeof(map));
20 for(i=0;i<N;++i)
21 cost[i] = INT_MAX;
22 memset(num_path,0,sizeof(num_path));
23 memset(num_res_gather,0,sizeof(num_res_gather));
24 memset(flag,0,sizeof(flag));
25 for (i=0;i<N;++i){
26 scanf("%d",&num_rescue[i]);
27 }
28 for (i=0;i<M;++i){
29 scanf("%d%d%d",&s,&e,&c);
30 map[s][e] = c;
31 map[e][s] = c;
32 }
33 cost[c1] = 0;
34 num_path[c1] = 1;
35 num_res_gather[c1] = num_rescue[c1];
36 flag[c1] = 1;
37 int path,rescue;
38 dijkstra(c1,c2,path,rescue);
39 printf("%d %d\n",path,rescue);
40 }
41 return 0;
42 }
43
44 int dijkstra(int start,int end,int& path,int& rescue)
45 {
46 int s = start;
47 while(!flag[end]){
48 int i;
49 for (i=0;i<N;++i){
50 if(!flag[i] && map[s][i]){
51 if (cost[s] + map[s][i] < cost[i]){
52 cost[i] = cost[s] + map[s][i];
53 num_path[i] = num_path[s];
54 num_res_gather[i] = num_res_gather[s] + num_rescue[i];
55 }
56 else if (!flag[i] && cost[s] + map[s][i] == cost[i]){
57 num_path[i] = num_path[s] + num_path[i];
58 num_res_gather[i] = ( num_res_gather[i] > num_res_gather[s] + num_rescue[i]?
59 num_res_gather[i] : num_res_gather[s]+num_rescue[i]);
60 }
61 }
62 }
63 i = 0;
64 while(flag[i++]);
65 int min_cost = cost[--i];
66 s = i;
67 for (++i;i<N;++i){
68 if (!flag[i] && cost[i] < min_cost){
69 min_cost = cost[i];
70 s = i;
71 }
72 }
73 flag[s] = 1;
74 }
75 path = num_path[end];
76 rescue = num_res_gather[end];
77 return cost[end];
78 }
