PAT 1003. Emergency (25)

题目地址： http://www.patest.cn/contests/pat-a-practise/1003

最短路

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 
 5 using namespace std;
 6 
 7 const int INF = 1000000000;
 8 const int MAX_NODES = 500 + 10;
 9 int edge[MAX_NODES][MAX_NODES];
10 int d[MAX_NODES];
11 int cnt[MAX_NODES];
12 int c_max[MAX_NODES];
13 int c[MAX_NODES];
14 bool flg[MAX_NODES];
15 int n, m, s, t;
16 
17 void shortest_path() {
18         memset(flg, false, sizeof(flg));
19         for (int i = 0; i < n; ++i) d[i] = i == s ? 0 : INF;
20         cnt[s] = 1;
21         c_max[s] = c[s];
22         for (int i = 0; i < n; ++i) {
23                 int x, mx = INF;
24                 for (int j = 0; j < n; ++j) {
25                         if (!flg[j] && d[j] <= mx) {
26                                 mx = d[x = j];
27                         }
28                 }
29                 flg[x] = true;
30                 for (int j = 0; j < n; ++j) {
31                         if (edge[x][j]) {
32                                 if (d[j] > d[x] + edge[x][j]) {
33                                         d[j] = d[x] + edge[x][j];
34                                         cnt[j] = cnt[x];
35                                         c_max[j] = c[j] + c_max[x];
36                                 } else if (d[j] == d[x] + edge[x][j]) {
37                                         cnt[j] += cnt[x];
38                                         c_max[j] = max(c[j] + c_max[x], c_max[j]);
39                                 }
40                         }
41                 }
42         }
43 }
44 int main() {
45         while(cin >> n >> m >> s >> t) {
46                 memset(edge, 0, sizeof(edge));
47                 for (int i = 0; i < n; ++i) cin >> c[i];
48                 for (int i = 0; i < m; ++i) {
49                         int c1, c2, l;
50                         cin >> c1 >> c2 >> l;
51                         edge[c1][c2] = l;
52                         edge[c2][c1] = l;
53                 }
54                 shortest_path();
55                 cout << cnt[t] << " " << c_max[t] << endl;
56         }
57         return 0;
58 }

 

posted @ 2015-04-13 13:33  ACSeed  Views(186)  Comments(0)    收藏  举报