POJ3268 Silver Cow Party —— 最短路

Posted 2020-10-08 h_z_cong

题目链接：http://poj.org/problem?id=3268

 

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24527		Accepted: 11164

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow‘s return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X 
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

Source

USACO 2007 February Silver

 

 

 

题解：

1.可知最短路分为两段， 各个点到X的最短距离，以及X到各个点的最短距离。

2.先求出X到各个点的最短距离， 记为dis1。

3.将各条边的方向取反， 然后再求出X到各个点的最短距离，记为dis2。因为边取反了，所以dis2实际是各个点到X的最短距离。

4.取dis1+dis2的最大值，即为答案。

 

 

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define rep(i,a,n) for(int (i) = a; (i)<=(n); (i)++)
13 #define ms(a,b) memset((a),(b),sizeof((a)))
14 using namespace std;
15 typedef long long LL;
16 const double EPS = 1e-8;
17 const int INF = 2e9;
18 const LL LNF = 9e18;
19 const int MOD = 1e9+7;
20 const int MAXN = 1e3+10;
21 
22 int n, m, X;
23 int g[MAXN][MAXN];
24 
25 int dis1[MAXN], dis2[MAXN];
26 bool vis[MAXN];
27 void dijkstra(int st, int dis[])
28 {
29     memset(vis, 0, sizeof(vis));
30     for(int i = 1; i<=n; i++)
31         dis[i] = (i==st?0:INF);
32 
33     for(int i = 1; i<=n; i++)
34     {
35         int k, minn = INF;
36         for(int j = 1; j<=n; j++)
37             if(!vis[j] && dis[j]<minn)
38                 minn = dis[k=j];
39 
40         vis[k] = 1;
41         for(int j = 1; j<=n; j++)
42             if(!vis[j] && g[k][j]!=INF)
43                 dis[j] = min(dis[j], dis[k]+g[k][j]);
44     }
45 }
46 
47 int main()
48 {
49     while(scanf("%d%d%d", &n, &m, &X)!=EOF)
50     {
51         for(int i = 1; i<=n; i++)
52             for(int j = 1; j<=n; j++)
53                 g[i][j] = INF;
54         for(int i = 1; i<=m; i++)
55         {
56             int u, v, w;
57             scanf("%d%d%d", &u, &v, &w);
58             g[u][v] = w;
59         }
60         dijkstra(X, dis1);  //第一次跑最短路，计算X到各点的最短距离
61 
62         for(int i = 1; i<=n; i++)   //将边取反
63             for(int j = i+1; j<=n; j++)
64                 swap(g[i][j], g[j][i]);
65         dijkstra(X, dis2);  //第二次跑最短路，计算X到各点的距离，但因为边取反了，所以实际上是各点到X的最短距离。
66 
67         int ans = 0;
68         for(int i = 1; i<=n; i++)   //取两段距离之和的最大值
69             ans = max(ans, dis1[i]+dis2[i]);
70         printf("%d\n", ans);
71     }
72 }

View Code