Bessie Come Home
Time Limit:JAVA/Others2000/1000MS Memory Limit:JAVA/Others131072/65536KB
Total Submit:6 Accepted:2
Description
It's dinner time, and the cows are out in their separate pastures. Farmer John rings the bell so they will start walking to the barn. Your job is to figure out which one cow gets to the barn first (the supplied test data will always have exactly one fastest cow).
Between milkings, each cow is located in her own pasture, though some pastures have no cows in them. Each pasture is connected by a path to one or more other pastures (potentially including itself). Sometimes, two (potentially self-same) pastures are connected by more than one path. One or more of the pastures has a path to the barn. Thus, all cows have a path to the barn and they always know the shortest path. Of course, cows can go either direction on a path and they all walk at the same speed.
The pastures are labeled `a'..`z' and `A'..`Y'. One cow is in each pasture labeled with a capital letter. No cow is in a pasture labeled with a lower case letter. The barn's label is `Z'; no cows are in the barn, though.
Input
| Line 1: |
Integer P (1 <= P <= 10000) the number of paths that interconnect the pastures (and the barn) |
| Line 2..P+1: |
Space separated, two letters and an integer: the names of the interconnected pastures/barn and the distance between them (1 <= distance <= 1000) |
Output
A single line containing two items: the capital letter name of the pasture of the cow that arrives first back at the barn, the length of the path followed by that cow.
Sample Input
5
A d 6
B d 3
C e 9
d Z 8
e Z 3
Sample Output
B 11
1
Floyd 算法：http://icpc.ahu.edu.cn:8080/AOJ/ 做的第一個(gè)圖論題
2圖的最短路徑問(wèn)題，到‘Z’的最短路徑；
3Floyd算法大概知道怎么用了 ，好像是動(dòng)態(tài)規(guī)劃實(shí)現(xiàn)的，不知道為什么這樣是對(duì)的
4O（N^3）求解最短路徑問(wèn)題，數(shù)據(jù)范圍超過(guò)400可能就危險(xiǎn)了
5#include<iostream>
6#include<string.h>
7using namespace std;
8int dis[53][53];
9const int INF=10000000;
10void Floyd(int n)
11
{
12
for(int k=1; k<=n; k++)
13 for(int i=1; i<=n; i++)
14 for(int j=1; j<=n; j++)
15 if(i!=k&&k!=j&&i!=j&&dis[i][k]+dis[k][j]<dis[i][j])
16 dis[i][j]=dis[i][k]+dis[k][j];
17
18
}
19
20
21int main()
22{
23 int p,i,j,k,d,n1,n2;
24 cin>>p;
25 memset(dis,0,sizeof (dis));
26 for(i=1; i<=52; i++)
27 for(j=1; j<=52; j++)
28 dis[i][j]=INF;
29
30 for(i=1; i<=p; i++)
31
{
32 char v1,v2;
33 cin>>v1>>v2>>d;
34 if(v1==v2)continue;
35 n1=(v1>='a'?v1-'a'+1:v1-'A'+26+1);
36 n2=(v2>='a'?v2-'a'+1:v2-'A'+26+1);
37 if(d<dis[n1][n2])dis[n1][n2]=dis[n2][n1]=d;
38
}
39
40 Floyd(52);
41
42 int min=INF+100;
43 char c;
44 for(i=27; i<=51; i++) //大寫(xiě)字母到Z
45 {
46 if(dis[i][52]<min){min=dis[i][52];c=i; }
47 }
48 cout<<char(c-27+'A')<<' '<<min<<endl;
49 //system("pause");
50 return 0;
51}
52