hdu 2066 一个人的旅行 (dij+heap)

Problem - 2066

　　入门SP，用dij+heap 0ms通过。

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 1 #define REP(i, n) for (int i = 0; i < (n); i++)
 2 #define REP_1(i, n) for (int i = 1; i <= (n); i++)
 3 typedef pair<LL, int> PLI;
 4 #define PB push_back
 5 #define MPR make_pair
 6 #define FI first
 7 #define SE second
 8 
 9 VPII rel[N];
10 LL dis[N];
11 multiset<PLI> HP;
12 
13 void input(int n, int s, int t) {
14     REP(i, 1002) rel[i].clear(), dis[i] = linf;
15     HP.clear();
16     int a, b, d;
17     REP(i, n) {
18         scanf("%d%d%d", &a, &b, &d);
19         rel[a].PB(MPR(b, d));
20         rel[b].PB(MPR(a, d));
21     }
22     REP(i, s) {
23         scanf("%d", &a);
24         rel[0].PB(MPR(a, 0));
25     }
26     REP(i, t) {
27         scanf("%d", &a);
28         rel[a].PB(MPR(1001, 0));
29     }
30 //    puts("OK!");
31 }
32 
33 LL dijHeap(int s, int t) {
34     dis[s] = 0ll;
35     HP.insert(MPR(0ll, s));
36     while (true) {
37         int id = HP.begin()->SE;
38 //        cout << "size " << SZ(HP) << endl;
39 //        cout << id << ' ' << HP.begin()->FI << ' ' << dis[id] << endl;
40         if (id == t) return HP.begin()->FI;
41         HP.erase(HP.begin());
42         REP(i, SZ(rel[id])) {
43             int x = rel[id][i].FI, d = rel[id][i].SE;
44             if (dis[x] > dis[id] + d) {
45 //                cout << x << ' ' << dis[x] << endl;
46                 if (dis[x] != linf) HP.erase(MPR(dis[x], x));
47                 HP.insert(MPR(dis[x] = dis[id] + d, x));
48             }
49         }
50     }
51 }
52 
53 int main() {
54 //    freopen("in", "r", stdin);
55     int n, s, t;
56     while (~scanf("%d%d%d", &n, &s, &t)) {
57         input(n, s, t);
58         cout << dijHeap(0, 1001) << endl;
59     }
60     return 0;
61 }

——written by Lyon

posted @ 2013-03-11 10:43 LyonLys 阅读(165) 评论(0) 收藏举报

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