POJ2387 Til the Cows Come Home 【Dijkstra】
题目链接:http://poj.org/problem?id=2387
题目大意;
题意:给出两个整数T,N,然后输入一些点直接的距离,求N和1之间的最短距离。。
#include <cstdio>
#include <cstring>
#define inf 0x3f3f3f3f
#define min(a,b) a<=b?a:b
int vis[], dis[]; //dis[j]表示起点到当前点的最短距离
int n, t;
int map[][]; void dij()
{
int i, j, cur, k;
memset(vis, , sizeof(vis));
for (i = ; i <= n; i++)(i == ) ? (dis[i] = ) : (dis[i] = inf);
cur = ;
for (i = ; i < n; i++) //循环n次,每次挑选没走过的到起点距离最短的点
{
vis[cur] = ;
for (j = ; j <= n; j++)
{
if (vis[j] == )
dis[j] = min(dis[j], map[cur][j] + dis[cur]); //更新每个没走过的点,到起点的最短距离
} //选择到起点距离最短的点
int g = inf; int x = ;
for (j = ; j <= n; j++)
{
if (dis[j] <= g && !vis[j])
{
g = dis[j];
x = j;
}
}
cur = x;
}
printf("%d\n", dis[n]); //输出终点到起点的最短距离
} int main()
{
int i, j, a, b, c;
while (scanf("%d%d", &t, &n) != EOF)
{
memset(map, inf, sizeof(map));
for (i = ; i <= t; i++)
{
scanf("%d%d%d", &a, &b, &c); //去除重边的情况
if (c < map[a][b])
map[a][b] = map[b][a] = c;
}
dij();
}
return ;
}
2018-04-01
POJ2387 Til the Cows Come Home 【Dijkstra】的更多相关文章
- POJ2387 Til the Cows Come Home【Kruscal】
题目链接>>> 题目大意: 谷仓之间有一些路径长度,然后要在这些谷仓之间建立一些互联网,花费的成本与长度成正比,,并且要使这些边连起来看的像一课“树”,然后使成本最大 解题思路: 最 ...
- POJ2387 Til the Cows Come Home(SPFA + dijkstra + BallemFord 模板)
Til the Cows Come Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37662 Accepted ...
- POj2387——Til the Cows Come Home——————【最短路】
A - Til the Cows Come Home Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & ...
- poj2387 Til the Cows Come Home 最短路径dijkstra算法
Description Bessie is out in the field and wants to get back to the barn to get as much sleep as pos ...
- poj2387 Til the Cows Come Home(Dijkstra)
题目链接 http://poj.org/problem?id=2387 题意 有n个路标,编号1~n,输入路标编号及路标之间相隔的距离,求从路标n到路标1的最短路径(由于是无向图,所以也就是求从路标1 ...
- POJ2387 Til the Cows Come Home (最短路 dijkstra)
AC代码 POJ2387 Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to ...
- (Dijkstra) POJ2387 Til the Cows Come Home
Til the Cows Come Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 81024 Accepted ...
- 怒学三算法 POJ 2387 Til the Cows Come Home (Bellman_Ford || Dijkstra || SPFA)
Til the Cows Come Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 33015 Accepted ...
- POJ 2387 Til the Cows Come Home 【最短路SPFA】
Til the Cows Come Home Description Bessie is out in the field and wants to get back to the barn to g ...
随机推荐
- gnuradio 创建cos_source
C++教程 ys_linux@computer:~$ gr_modtool nm kcd Creating out-of-tree module in ./gr-kcd... Done. Use 'g ...
- python(2): If/for/函数/try异常/调试/格式输出%
(一) if if a1==a2: print('ok') if: else: if: elif: ... else: 注意缩进 猜数字游戏 from random import randint ...
- Vuex状态管理模式的面试题及答案
转载:点击查看原文 1.vuex有哪几种属性? 答:有五种,分别是 State. Getter.Mutation .Action. Module 2.vuex的State特性是? 答: 一.Vuex就 ...
- ajax之阴影效果实现(对象函数方法)
shadow.js文件内容jQuery.fn.shadow = function () { //获取到每个已封装的元素 //this表示jQuery对象 this.each(function () { ...
- C++ Primer 笔记——变量
1. 初始化不是赋值,初始化的含义是创建变量时赋予其一个初始值,而赋值的含义是把对象的当前值擦除,而以一个新值来代替. 2.使用列表初始化内置类型的变量时,如果初始值存在丢失信息的风险,则编译器将报错 ...
- 重建控制文件报错 ORA-01503 ORA-01192
1. 错误信息 ORA-: CREATE CONTROLFILE failed ORA-: must have at least one enabled thread 2. 重建脚本 CREATE C ...
- Android 中的缓存
AsimpleCache 1.它可以缓存什么东西? 普通的字符串. json. 序列化的java对象 字节数字. 2.主要特色 1:轻,轻到只有一个JAVA文件. 2:可配置,可以配置缓存路径,缓存 ...
- Ubuntu 进入、退出命令行的快捷键
进入: Ctrl+Alt+F1 退出: Ctrl+Alt+F7(或者 Alt+F7) 进入命令行窗口:Ctrl+Alt+T
- Stuck on "Authenticating with iTunes Store"
https://forums.developer.apple.com/thread/76803 Try this, it fixed it for me. Open Terminal and run: ...
- Visual Studio上编译ncnn
prerequisite 是为了在PC上熟悉ncnn的基本代码,所以用Visual Studio来配置的. 期间用过VS2013(update5)/VS2015/VS2017,反正都是基于CMake生 ...