[kuangbin]带你飞之'最短路练习'专题(未完成)

 1 /*
 2  * @Promlem: 
 3  * @Time Limit: ms
 4  * @Memory Limit: k
 5  * @Author: pupil-XJ
 6  * @Date: 2019-10-25 08:14:08
 7  * @LastEditTime: 2019-10-25 08:25:39
 8  */
 9 #include<cstdio>
10 #include<cstring>
11 #include<cmath>
12 #include<iostream>
13 #include<string>
14 #include<algorithm>
15 #include<vector>
16 #include<queue>
17 #include<stack>
18 #include<set>
19 #include<map>
20 #define rep(i, n) for(int i=0;i!=n;++i)
21 #define per(i, n) for(int i=n-1;i>=0;--i)
22 #define Rep(i, sta, n) for(int i=sta;i!=n;++i)
23 #define rep1(i, n) for(int i=1;i<=n;++i)
24 #define per1(i, n) for(int i=n;i>=1;--i)
25 #define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
26 #define L k<<1
27 #define R k<<1|1
28 #define mid (tree[k].l+tree[k].r)>>1
29 using namespace std;
30 const int INF = 0x3f3f3f3f;
31 typedef long long ll;
32 
33 inline int read() {
34     char c=getchar();int x=0,f=1;
35     while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
36     while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
37     return x*f;
38 }
39 // -----------------------------------------------------
40 const int MAXN = 1000+5;
41 int T, n;
42 int G[MAXN][MAXN];
43 
44 int vis[MAXN], dis[MAXN];
45 
46 int dij() {
47     rep1(i, n) {
48         dis[i] = G[i][1];
49         vis[i] = 0;
50     }
51     vis[1] = 1;
52     rep(i, n-1) {
53         int minn = INF;
54         int t;
55         rep1(j, n) {
56             if(!vis[j] && dis[j] < minn) {
57                 minn = dis[j];
58                 t = j;
59             }
60         }
61         vis[t] = 1;
62         rep1(j, n) {
63             if(!vis[j] && dis[j] > dis[t] + G[t][j]) {
64                 dis[j] = dis[t] + G[t][j];
65             }
66         }
67     }
68     return dis[n];
69 }
70 
71 int main() {
72     ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
73     int s, e, v;
74     while(cin >> T >> n) {
75         rep1(i, n) rep1(j, n) G[i][j] = INF;
76         rep(i, T) {
77             cin >> s >> e >> v;
78             if(v < G[s][e]) G[s][e] = G[e][s] = v;
79         }
80         cout << dij() << "\n";
81     }
82     return 0;
83 }