poj 2387 Dijkstra裸体 有重边陷阱
2013-04-09 22:18
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Til the Cows Come Home
Description
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.
Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various
lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
Input
* Line 1: Two integers: T and N
* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
Output
* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.
Sample Input
Sample Output
Hint
INPUT DETAILS:
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 22964 | Accepted: 7706 |
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.
Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various
lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
Input
* Line 1: Two integers: T and N
* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
Output
* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.
Sample Input
5 5 1 2 20 2 3 30 3 4 20 4 5 20 1 5 100
Sample Output
90
Hint
INPUT DETAILS:
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <string> #include <map> #include <set> #include <vector> #include <iomanip> #include <queue> #include <climits> using namespace std; const int MAXN = 1005; const int INF = INT_MAX; // a very big number int n, t; int g[MAXN][MAXN]; // 2 dimension array used to store the graph int dis[MAXN]; void load() { cin >> t >> n; // do the initialization for (int i = 2; i <= n; ++i) dis[i] = INF; for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) g[i][j] = INF; // load the data for (int i = 1; i <= t; ++i) { int a, b, c; cin >> a >> b >> c; if (g[a][b] > c) // there might be duplicate edges, wo choose the shortest one g[a][b] = g[b][a] = c; } } bool used[MAXN]; void work() { while (true) { // select the nearest vertex that hasn't been used yet int minu = -1; int u; for (u = 1; u <= n; ++u) { if (!used[u]) { if (-1 == minu || dis[u] < dis[minu]) { minu = u; } } } u = minu; if (u == n) // we have got the destination break; used[u] = true; // do the relax operation for (int v = 1; v <= n; ++v) if (g[u][v] != INF) { if (dis[v] > dis[u]+g[u][v]) { dis[v] = dis[u]+g[u][v]; } } } cout << dis << endl; } int main() { load(); work(); return 0; }
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