POJ 3268 Silver Cow Party (最短路dijkstra)
Silver Cow Party
题目链接:
http://acm.hust.edu.cn/vjudge/contest/122685#problem/D
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
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.
##题意:
n个人分别住在1~n个点,他们要去某个点聚会并且结束后返回家中.
求每个人的最短路程. (往返路径可以不一样).
##题解:
先用dijkstra求出终点到所有点的最短路径(返程).
再沿着反向路径跑一次dijkstra,求出所有点到终点的最短路径.
##代码:
``` cpp
#include
#include
#include
#include
#include
#define mid(a,b) ((a+b)>>1)
#define LL long long
#define maxn 1010
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;
int n,m;
int value[maxn][maxn];
int dis[maxn];
bool vis[maxn];
int dis2[maxn];
void dijkstra(int s) {
memset(vis, 0, sizeof(vis));
for(int i=1; i<=n; i++) dis[i] = inf;
dis[s] = 0;
for(int i=1; i<=n; i++) {
int p, mindis = inf;
for(int j=1; j<=n; j++) {
if(!vis[j] && dis[j]<mindis)
mindis = dis[p=j];
}
vis[p] = 1;
for(int j=1; j<=n; j++) {
if(dis[j] > dis[p]+value[p][j]) {
dis[j] = dis[p] + value[p][j];
}
}
}
}
void dijkstra2(int s) {
memset(vis, 0, sizeof(vis));
for(int i=1; i<=n; i++) dis2[i] = inf;
dis2[s] = 0;
for(int i=1; i<=n; i++) {
int p, mindis = inf;
for(int j=1; j<=n; j++) {
if(!vis[j] && dis2[j]<mindis)
mindis = dis2[p=j];
}
vis[p] = 1;
for(int j=1; j<=n; j++) {
if(dis2[j] > dis2[p]+value[j][p]) {
dis2[j] = dis2[p] + value[j][p];
}
}
}
}
int main(int argc, char const *argv[])
{
//IN;
int aim;
while(scanf("%d %d %d", &n,&m,&aim) != EOF)
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
value[i][j] = inf;
while(m--){
int u,v,w; cin>>u>>v>>w;
if(w < value[u][v]) value[u][v] = w;
}
dijkstra(aim);
dijkstra2(aim);
int ans = 0;
for(int i=1; i<=n; i++) if(dis[i]!=inf && dis2[i]!=inf)
ans = max(ans, dis[i]+dis2[i]);
printf("%d\n", ans);
}
return 0;
}
POJ 3268 Silver Cow Party (最短路dijkstra)的更多相关文章
- POJ 3268 Silver Cow Party 最短路—dijkstra算法的优化。
POJ 3268 Silver Cow Party Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbe ...
- poj 3268 Silver Cow Party(最短路dijkstra)
描述: One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the bi ...
- POJ 3268 Silver Cow Party (双向dijkstra)
题目链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS Memory Limit: 65536K Total ...
- POJ 3268 Silver Cow Party 最短路
原题链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS Memory Limit: 65536K Total ...
- poj 3268 Silver Cow Party (最短路算法的变换使用 【有向图的最短路应用】 )
Silver Cow Party Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13611 Accepted: 6138 ...
- POJ 3268 Silver Cow Party (最短路径)
POJ 3268 Silver Cow Party (最短路径) Description One cow from each of N farms (1 ≤ N ≤ 1000) convenientl ...
- POJ 3268——Silver Cow Party——————【最短路、Dijkstra、反向建图】
Silver Cow Party Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Su ...
- DIjkstra(反向边) POJ 3268 Silver Cow Party || POJ 1511 Invitation Cards
题目传送门 1 2 题意:有向图,所有点先走到x点,在从x点返回,问其中最大的某点最短路程 分析:对图正反都跑一次最短路,开两个数组记录x到其余点的距离,这样就能求出来的最短路以及回去的最短路. PO ...
- POJ 3268 Silver Cow Party 单向最短路
Silver Cow Party Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 22864 Accepted: 1044 ...
随机推荐
- Oracle 数据集成的实际解决方案
就针对市场与企业的发展的需求,Oracle公司提供了一个相对统一的关于企业级的实时数据解决方案,即Oracle数据集成的解决方案.以下的文章主要是对其解决方案的具体描述,望你会有所收获. Oracle ...
- Sublime Text汉化方法和注册码
汉化方法 安装 SublimeText3 汉化包运行SublimeText3 点击 Preferneces -> Browse Packages 会打开 X:\..\Sublime Text 3 ...
- CSS+DIV问题!DIV的最小高度问题!
DIV层的最小高度问题!就是一个DIV有个最小高度,但是如果DIV层中的内容很多,DIV的高度会根据内容而进行拉长!要求IE6.IE7还有firefox都要兼容!我试了很多网上的方法都不好用!请测试后 ...
- UVa 12661 (单源最短路) Funny Car Racing
题意: 有一个赛车跑道,可以看做一个加权有向图.每个跑道(有向边)还有一个特点就是,会周期性地打开a秒,然后关闭b秒.只有在赛车进入一直到出来,该跑道一直处于打开状态,赛车才能通过. 开始时所有跑道处 ...
- 纯CSS3带动画效果导航菜单
随着互联网的发展,网页能表现的东西越来越多.由最开始单纯的文字和链接构成的网页,到后来的表格布局,再到div+css模式,现在发展到了html+css3.网页能表达的东西越来越多,css3兴起已经很多 ...
- 【JSP】JSP动态显示时间
function showtime() { var today; var hour; var second; var minute; var year; var month; var date; va ...
- 【转】linux驱动开发的经典书籍
原文网址:http://www.cnblogs.com/xmphoenix/archive/2012/03/27/2420044.html Linux驱动学习的最大困惑在于书籍的缺乏,市面上最常见的书 ...
- 连接Excel时出现未指定的错误
使用 strConn = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" + filepath + ";Extended ...
- Spring AOP (上)
工作忙,时间紧,不过事情再多,学习是必须的.记得以前的部门老大说过:“开发人员不可能一天到晚只有工作,肯定是需要自我学习.第一:为了更充实自己,保持进步状态.第二:为了提升技术,提高开发能力.第三:保 ...
- 正在连接...ORA-12541: TNS: 无监听程序
明明已经在Net Configuration Assistant中配置过监听程序并启动过.但在测试本地网络服务名配置中扔提示以上错误"正在连接...ORA-12541: TNS: 无监听程序 ...