uestc 方老师的分身 II

题意：求s到t走过边数大于k的最短路

思路：邻接表实现，用w[u][e]来维护(看的大牛博客),u表示当前点，e表示已经经过多少条边。感觉有点类似DP。

在边数大于k的处理上，发现还是使之等于k（K<=50），节省存储空间。

spfa算法实现。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<string>
 6 #include<queue>
 7 #include<algorithm>
 8 #include<map>
 9 #include<iomanip>
10 #include<climits>
11 #include<string.h>
12 #include<cmath>
13 #include<stdlib.h>
14 #include<vector>
15 #define INF 1000000007
16 #define MAXN 5010
17 #define maxn 1000010
18 #define Mod 1000007
19 #define N 1010
20 using namespace std;
21 typedef long long LL;
22
23 struct node{
24     int u, e;
25 };
26 int n, m, u, v, wt;
27 int s, t, k;
28 int ans;
29 int dist[MAXN][55];
30 bool inq[MAXN][55];
31 vector< pair<int, int> > G[MAXN];
32
33 void spfa(int s)
34 {
35     int u, v, e, i, j, len;
36     for (i = 0; i <= n; ++i)
37         for (j = 0; j <= 50; ++j) {
38             dist[i][j] = INF;
39             inq[i][j] = 0;
40         }
41     queue<node> q;
42     ans = INF;
43     dist[s][0] = 0;
44     inq[s][0] = true;
45     q.push({s,0});
46     while (!q.empty()) {
47         node tmp = q.front();
48         q.pop();
49         u = tmp.u;
50         e = tmp.e;
51         if (u == t && e == k)
52             ans = min(dist[u][e],ans);
53         inq[u][e] = false;
54         int time;
55         if (e + 1 > k) time = k;
56         else time = e + 1;
57         for (i = 0; i < G[u].size(); ++i) {
58             v = G[u][i].first;
59             len = G[u][i].second;
60             if (dist[v][time] > len + dist[u][e]) {
61                 dist[v][time] = len + dist[u][e];
62                 if (!inq[v][time]) {
63                     q.push({ v, time });
64                     inq[v][time] = true;
65                 }
66             }
67         }
68     }
69 }
70
71 int main()
72 {
73     while (cin >> n >> m) {
74         for (int i = 0; i < MAXN; ++i) G[i].clear();
75         for (int i = 0; i < m; ++i) {
76             cin >> u >> v >> wt;
77             G[u].push_back({ v, wt });
78             G[v].push_back({ u, wt });
79         }
80         cin >> s >> t >> k;
81         spfa(s);
82         if (ans == INF) puts("-1");
83         else cout << ans << endl;
84     }
85     return 0;
86 }