POJ 2387 Til the Cows Come Home(Dijkstra算法)
2018-04-11 04:59
477 查看
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
5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100
Sample Output
90
题目大意:
给你一个数t,代表有几条已知的道路,再给出一个数n ,代表有n个点,让你求出从1到n的最短距离为多少
解题思路
标准迪杰斯特拉算法题目
算法思路过程如下:
参考代码
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #define INF 0x3f3f3f3f #define MME(a) memset(a,0,sizeof(a)) using namespace std; int book[1005];//记录改点是否走过 int dp[1005];//记录1到任意节点的距离 int path[1005][1005];//记录两点之间的距离 int main() { int t, n; while (~scanf("%d %d", &t, &n)) { MME(book); MME(dp); MME(path); for (int i = 1;i <= n;i++) { for (int j = 1;j <= n;j++) { if (i == j)//由于不能自环,所以主对角线置零 path[i][j] = 0; else path[i][j] = INF; } } for (int i = 1;i <= t;i++) { int st, ed, len; scanf("%d %d %d", &st, &ed, &len); /*坑点:需要考虑两点重复输入更新*/ path[st][ed] = path[ed][st] = min(path[st][ed], len); } for (int i = 1;i <= n;i++) { dp[i] = path[1][i];/*dp数组存储1到任意节点的距离,如果没有关系就是正无穷*/ } book[1] = 1; int pos = 0; for (int i = 1;i <= n-1;i++) { int min = INF; for (int j = 1;j <= n;j++) { /*如果改点没走过,且1到该节点为最短距离时*/ if (!book[j] && dp[j] < min) { min = dp[j]; pos = j; } } book[pos] = 1; /*说明1到该节点的距离已为最短,然后标记上*/ for (int k = 1;k <= n;k++) { if (path[pos][k] != INF)/*如果pos到k有关系*/ { /*判断1直接到k点和1点到pos点再由pos点到k点谁距离短,然后赋值*/ if (dp[k] > dp[pos] + path[pos][k]) dp[k] = dp[pos] + path[pos][k]; } } } printf("%d\n", dp ); } }
相关文章推荐
- Til the Cows Come Home(poj 2387 Dijkstra算法)
- POJ 2387 Til the Cows Come Home (最短路径,Dijkstra算法)
- poj 2387 Til the Cows Come Home(dijkstra算法)
- POJ 2387 Til the Cows Come Home(模板——Dijkstra算法)
- (阶段三 dijkstra算法温习 1.6)POJ 2387 Til the Cows Come Home(使用dijkstra算法求单源起点和单源终点的最短路径)
- POJ-2387 Til the Cows Come Home(最短路 Dijkstra算法)
- POJ 2387 Til the Cows Come Home(最短路,Dijkstra算法)
- POJ 2387 Til the Cows Come Home(Dijkstra算法)
- POJ 2387 Til the Cows Come Home dijkstra算法 用邻接表和邻接矩阵
- poj 2387 Til the Cows Come Home(dijkstra算法)
- Til the Cows Come Home(poj 2387 Dijkstra算法(单源最短路径))
- poj 2387 Til the Cows Come Home -- 最短路dijstra
- 2387 poj Til the Cows Come Home【dijkstra,经典&&基础】
- Til the Cows Come Home POJ - 2387 单源最短路 SPFA实现
- POJ 2387 Til the Cows Come Home --最短路模板题
- POJ 2387 Til the Cows Come Home 最短路 Dijstra
- POJ 2387 Til the Cows Come Home --单源最短路径
- POJ 2387-Til the Cows Come Home(Dijkstra+堆优化)
- POJ 2387 Til the Cows Come Home(Dijkstra简单题)
- poj 2387 Til the Cows Come Home (最短路)