| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5720 | Accepted: 2266 |
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
1邻接矩阵A的K次方C=A^K,C[i][j]表示i点到j点正好经过K条边的路径数
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
int n,t,s,e,res;
int hash[1010],dist[205][205],temp[205][505],f[205][205];
void floyd()
{
while(n)
{
if(n&1)
{
memset(temp,0x3f,sizeof(temp));
for(int k=1;k<=res;k++) for(int i=1;i<=res;i++)
for(int j=1;j<=res;j++) temp[i][j]=min(temp[i][j],f[i][k]+dist[k][j]);
for(int i=1;i<=res;i++) for(int j=1;j<=res;j++) f[i][j]=temp[i][j];
}
memset(temp,0x3f,sizeof(temp));
for(int k=1;k<=res;k++) for(int i=1;i<=res;i++)
for(int j=1;j<=res;j++) temp[i][j]=min(temp[i][j],dist[i][k]+dist[k][j]);
for(int i=1;i<=res;i++) for(int j=1;j<=res;j++) dist[i][j]=temp[i][j];
n>>=1;
}
return ;
}
int main()
{
int z,x,y;
while(scanf("%d%d%d%d",&n,&t,&s,&e)!=EOF)
{
res=0;
memset(dist,0x3f,sizeof(dist));
memset(f,0x3f,sizeof(f));
for(int i=0;i<=200;i++) f[i][i]=0;
memset(hash,0,sizeof(hash));
for(int i=0;i<t;i++)
{
scanf("%d%d%d",&z,&x,&y);
if(!hash[x]) hash[x]=++res;
if(!hash[y]) hash[y]=++res;
dist[hash[x]][hash[y]]=dist[hash[y]][hash[x]]=z;
}
floyd();
printf("%d\n",f[hash[s]][hash[e]]);
}
return 0;
}
原文：http://www.cnblogs.com/water-full/p/4486701.html