POJ 2387 dijkstra水题
2014-07-03 09:53
225 查看
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.
#include <iostream>
#include <cstdio>
#include <fstream>
#include <cstring>
using namespace std;
int map[1005][1005];
int dijkstra(int n)
{
int answer[1005];
int flag[1005],min,index;
memset(flag,0,sizeof(flag));
for(int i=1;i<=n;i++)answer[i]=map
[i];
answer
=0;
flag
=1;
for(int i=1;i<=n-1;i++)
{
min=9999999;
for(int j=1;j<=n;j++)
{
if(answer[j]<min&&flag[j]==0){min=answer[j];index=j;}
}
flag[index]=1;
for(int j=1;j<=n;j++)
{
if(map[index][j]<9999999&&flag[j]==0&&answer[j]>answer[index]+map[index][j])answer[j]=answer[index]+map[index][j];
}
}
return answer[1];
}
int main()
{
int T,N,city1,city2,minute;//N is points <,T is ways
cin>>T>>N;
for(int i=1;i<=N;i++)
{
for(int j=1;j<=N;j++)map[i][j]=9999999;
map[i][i]=0;
}
for(int i=1;i<=T;i++)
{
cin>>city1>>city2>>minute;
if(map[city1][city2]>minute)map[city1][city2]=map[city2][city1]=minute;
}
cout<<dijkstra(N);
}
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 28742 | Accepted: 9667 |
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 <fstream>
#include <cstring>
using namespace std;
int map[1005][1005];
int dijkstra(int n)
{
int answer[1005];
int flag[1005],min,index;
memset(flag,0,sizeof(flag));
for(int i=1;i<=n;i++)answer[i]=map
[i];
answer
=0;
flag
=1;
for(int i=1;i<=n-1;i++)
{
min=9999999;
for(int j=1;j<=n;j++)
{
if(answer[j]<min&&flag[j]==0){min=answer[j];index=j;}
}
flag[index]=1;
for(int j=1;j<=n;j++)
{
if(map[index][j]<9999999&&flag[j]==0&&answer[j]>answer[index]+map[index][j])answer[j]=answer[index]+map[index][j];
}
}
return answer[1];
}
int main()
{
int T,N,city1,city2,minute;//N is points <,T is ways
cin>>T>>N;
for(int i=1;i<=N;i++)
{
for(int j=1;j<=N;j++)map[i][j]=9999999;
map[i][i]=0;
}
for(int i=1;i<=T;i++)
{
cin>>city1>>city2>>minute;
if(map[city1][city2]>minute)map[city1][city2]=map[city2][city1]=minute;
}
cout<<dijkstra(N);
}
相关文章推荐
- Dijkstra单源最短路径,POJ(2387)
- (简单) POJ 2387 Til the Cows Come Home,Dijkstra。
- poj 2387 Til the Cows Come Home dijkstra
- POJ 2387 Til the Cows Come Home(Dijkstra)
- POJ - 2387 Til the Cows Come Home(Dijkstra)
- POJ 2387 Til the Cows Come Home(Dijkstra判重边)
- POJ 2387 Til the Cows Come Home(Dijkstra)
- POJ 1502 Dijkstra最短路水题
- POJ ~ 2387 ~ Til the Cows Come Home(Dijkstra)
- Dijkstra 优先级队列 实现 POJ 2387
- 基本算法dijkstra的POJ水题推荐
- POJ 2387 第一道Dijkstra AC的很纠结
- poj 2387(Dijkstra优先队列优化)
- POJ 2387 Til the Cows Come Home - (Dijkstra)
- poj 2387 Dijkstra裸体 有重边陷阱
- POJ 2387 - Til the Cows Come Home ( dijkstra求最短路 )
- POJ - 2387 Til the Cows Come Home(Dijkstra SPFA 邻接矩阵 邻接表)
- 2387 poj Til the Cows Come Home【dijkstra,经典&&基础】
- poj 2387 Til the Cows Come Home (最短路,dijkstra模版题)
- POJ 1125 Stockbroker Grapevine 最短路水题floyd || dijkstra