Poj 2387 单源最短路径

Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 48813		Accepted: 16604

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

Hint

INPUT DETAILS: 

There are five landmarks. 

OUTPUT DETAILS: 

Bessie can get home by following trails 4, 3, 2, and 1.

单元最短路径的题，DP解决。

先输入的是边，然后输入的是点，由于可以往复的走，所以建立数组要建立n*n的来记录往返，由于往返路径相同，所以只需要用  a[i][j]=a[j][i]即可。先把所有数值赋值为最大，最大值代表着不通，用数组v来标记走没走，用数组dp来记录路程，a来储存可走路径。详解见代码。

另注意，a[i][j]若有特殊值，代表着从i到j是通路的。

#include<stdio.h>
#include<string.h>
int max=1000000;        //设置一个最大值 
int dp[1005],v[1005];   //dp储存路程，v用来标记 
int a[1005][1005];      //储存可走路径 
int T,N;                
void DP()
{
int i,j,k,min;
memset(v,0,sizeof(v));        //将v设为0，代表没走到 
v[1]=1;                       //从1开始，默认1已经走了 
for(i=1;i<N;i++)              //遍历从第i个能走到第几个且记录最短路径 
{
k=1;                      //记录 目前走到的点 
min=10000000;             //每循环一次，min重新赋值 
for(j=1;j<=N;j++)
{
if(!v[j]&&min>dp[j])  //j没被标记且到j到原点这条路是通的 
{
min=dp[j];       //min代表从第一个点到第j个点的最短路径 
k=j;             //k记录从第一个点到第n个点 
}
}
v[k]=1;                  //走到，被标记 
for(j=1;j<=N;j++)        // 
{
if(!v[j]&&dp[j]>dp[k]+a[k][j])  //没被标记且1到j的路程大于1到k加上 k到j的 
dp[j]=dp[k]+a[k][j];            //1到j的路程等于1到k加上k到j的 
}
}
printf("%d\n",dp[N]);                  //1到N的路程 
}


int main()
{
int i,j,k,x,y,c;
while(scanf("%d%d",&T,&N)!=EOF)
{
for(i=1;i<=N;i++)
        {
            a[i][i]=0;          //对角线上的点代表着原地不动 
            for(j=1;j<i;j++)
            a[i][j]=a[j][i]=max;
        }
for(i=1;i<=T;i++)
{
scanf("%d%d%d",&x,&y,&c); 
if(c<a[x][y])       //若通路，则记录 
a[x][y]=a[y][x]=c;
}
for(i=1;i<=N;i++)
dp[i]=a[1][i];          //用d来储存各数到第一个点的距离 
DP();
}
return 0;
}