最短路 Codeforces Round #103 (Div. 2) D. Missile Silos

SPFA算法详解

最新推荐文章于 2022-12-16 16:19:21 发布

转载最新推荐文章于 2022-12-16 16:19:21 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/Running-Time/p/4659880.html

题目传送门

 1 /*
 2     最短路： 不仅扫描边，还要扫描点；点有两种情况，一种刚好在中点，即从u，v都一样，那么最后/2
 3                 还有一种是从u，v不一样，两种的距离都是l
 4     模板错了，逗了好久:(
 5 */
 6 #include <cstdio>
 7 #include <algorithm>
 8 #include <cstring>
 9 #include <cmath>
10 #include <vector>
11 #include <queue>
12 using namespace std;
13 
14 const int MAXN = 1e5 + 10;
15 const int INF = 0x3f3f3f3f;
16 int d[MAXN];
17 int cnt[MAXN];
18 bool vis[MAXN];
19 vector<pair<int, int> > G[MAXN];
20 int n, m, s, l, ans, ans2;
21 
22 void SPFA(void) {
23     memset (vis, false, sizeof (vis));
24     memset (d, INF, sizeof (d));  d[s] = 0; 
25     queue<int> Q;   Q.push (s);
26     while (!Q.empty ()) {
27         int u = Q.front (); Q.pop ();
28         vis[u] = false;
29         for (int i=0; i<G[u].size (); ++i)  {
30             int v = G[u][i].first;    int w = G[u][i].second;
31             if (d[v] > d[u] + w)    {
32                 d[v] = d[u] + w;
33                 if (!vis[v])    {
34                     vis[v] = true;  Q.push (v);
35                 }
36             }    
37         }
38     }
39 }
40 
41 int main(void)  {       //Codeforces Round #103 (Div. 2) D. Missile Silos
42     //freopen ("spfa.in", "r", stdin);
43     
44     while (scanf ("%d%d%d", &n, &m, &s) == 3)   {
45         for (int i=1; i<=m; ++i)    {
46             int u, v, w;    scanf ("%d%d%d", &u, &v, &w);
47             G[u].push_back (make_pair (v, w)); G[v].push_back (make_pair (u, w));
48         }
49         scanf ("%d", &l);   SPFA ();
50         
51         ans = ans2 = 0;
52         for (int i=1; i<=n; ++i)    {
53             for (int j=0; j<G[i].size (); ++j)  {
54                 int u = i, v = G[i][j].first, w = G[i][j].second;
55                 if (d[u] < l && l - d[u] < w)   {
56                     if (w - (l-d[u]) + d[v] > l)    ans++;
57                     else if (d[u] + d[v] + w == 2 * l)  ans2++;
58                 }
59             }
60             if (d[i] == l)  ans++;
61         }
62         printf ("%d\n", ans + ans2 / 2);
63     }
64 
65     return 0;
66 }

转载于:https://www.cnblogs.com/Running-Time/p/4659880.html