# POJ 3268 Silver Cow Party C++版

POJ题解 3273浏览

Dijkstra(单源最短路径)算法的应用

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 Ti (1 ≤ Ti ≤ 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: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti 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

Slyar:题目大意是给出一个有向图的带权邻接矩阵，然后给定目标节点X，求出其余结点出发到结点X再从X回到出发结点的最短路径和中最大的那个…=_=是不是很晕？晕的话去看英文…

```#include

using namespace std;

const int N = 1010;
const int MAX = 1000001;
int n, m;
int map1[N][N];
int map2[N][N];
int dist1[N];
int dist2[N];

// dist[]记录节点x到其余各点的最短路径长度
void dijkstra(int map[][N], int dist[], int x)
{
int u, min;
// 记录节点是否求解完毕
int visited[N] = {0};
// 节点x到自己的距离为0
dist[x] = 0;
// 求解其余n-1个节点到x的最短路径
for (int i = 1; i <= n; i++)
{
// 选择与当前节点相邻且未访问的节点中距离最短的节点u
for (int j = 1; j <= n; j++)
{
if (!visited[j] && dist[j] < min)
{
min = dist[j];
u = j;
}
}
// 标记u已求解
visited[u] = 1;
// 根据u来更新其余节点
for (int j = 1; j <= n; j++)
{
if (!visited[j] && dist[j] > dist[u] + map[u][j])
{
dist[j] = dist[u] + map[u][j];
}
}
}
}

int main()
{
int ai, bi, ti, x;
int ans = 0;
cin >> n >> m >> x;
//初始化
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
map1[i][j] = MAX;
map2[i][j] = MAX;
}
dist1[i] = MAX;
dist2[i] = MAX;
}
//读入矩阵
for (int i = 1; i <= m; i++)
{
cin >> ai >> bi >> ti;
map1[ai][bi] = ti;
map2[bi][ai] = ti;
}

dijkstra(map1, dist1, x);
dijkstra(map2, dist2, x);

for (int i = 1; i <= n; i++)
{
if (ans < dist1[i] + dist2[i])
{
ans = dist1[i] + dist2[i];
}
}

cout << ans << endl;
return 0;
}
```

### 网友最新评论 (1)

1. 楼主是不是写错了，样例都过不了。。。。。。。。。。
向伟6年前 (2015-07-31)回复