| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5411 | Accepted: 2153 |
Description
For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.
Each trail connects two different intersections (1 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the lengthi of each trail (1 ≤ lengthi ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.
To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.
Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.
Input
* Line 1: Four space-separated integers: N, T, S, and E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: lengthi , I1i , and I2i
Output
* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.
Sample Input
2 6 6 4 11 4 6 4 4 8 8 4 9 6 6 8 2 6 9 3 8 9
Sample Output
10
n次floyd，原来flody也可以像矩阵一样快速幂。具体的可以看论文《矩阵乘法在信息学的应用》
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int inf=(1<<30)-1;
int n;
int hash[3010];//映射
struct matrix
{
int ma[210][210];
matrix()
{
for(int i=0;i<210;i++)
for(int j=0;j<210;j++)
ma[i][j]=inf;
}
};
matrix floyd(matrix x,matrix y)
{
matrix ans;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
for(int k=1;k<=n;k++)
ans.ma[i][j]=min(ans.ma[i][j],x.ma[i][k]+y.ma[k][j]);
}
}
return ans;
}
matrix pow(matrix a,int m)
{
matrix ans;
for(int i=1;i<=n;i++)
ans.ma[i][i]=0;
while(m)
{
if(m&1)
ans=floyd(ans,a);
a=floyd(a,a);
m=m>>1;
}
return ans;
}
int main()
{
int k,t,s,e;
while(~scanf("%d%d%d%d",&k,&t,&s,&e))
{
int x,y,d;
memset(hash,0,sizeof(hash));
n=1;
matrix a;
for(int i=0;i<t;i++)
{
scanf("%d%d%d",&d,&x,&y);
if(!hash[x])
hash[x]=n++;
if(!hash[y])
hash[y]=n++;
a.ma[hash[x]][hash[y]]=a.ma[hash[y]][hash[x]]=d;
}
n=n-1;
matrix ans;
ans=pow(a,k);
printf("%d\n",ans.ma[hash[s]][hash[e]]);
}
return 0;
}
原文地址：http://blog.csdn.net/caduca/article/details/40615207
