poj 3613 Cow Relays(행렬 연결 도론에서의 응용)

Cow Relays
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 4003
 
Accepted: 1573
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

Source
USACO 2007 November Gold
제목:http://poj.org/problem?id=3613
제목: 그림 한 장 드릴게요. 변수가 n인 s에서 e까지의 가장 짧은 길을 구해 주세요.
분석: n이 비교적 커서 처음에는 별 생각이 없었는데 나중에 갑자기 하나의 행렬로 점 간의 전환 관계를 나타낸 다음에 행렬로 연승하는 것을 생각했다...정해랑 좀 가까웠는데 반응이 안 나왔어요.
사실 행렬의 뜻을 몇 개의 테두리를 사용한 후 i에서 j까지의 최단 거리를 누르면 된다...매번 매트릭스 곱하기는floyd 방식으로 이동합니다...
코드:

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef __int64 LL;
const int mm=1111;
const int mn=222;
const LL oo=1e15;
int id[mm];
LL ans[mn][mn],tmp[mn][mn],dis[mn][mn];
int n,T,m,s,t;
void prepare(LL a[mn][mn])
{
    for(int i=0;i<mn;++i)
        for(int j=0;j<mn;++j)
            a[i][j]=oo;
}
void Multi(LL a[mn][mn],LL b[mn][mn])
{
    int i,j,k;
    prepare(tmp);
    for(k=0;k<m;++k)
        for(i=0;i<m;++i)
            for(j=0;j<m;++j)
                tmp[i][j]=min(tmp[i][j],a[i][k]+b[k][j]);
    for(i=0;i<m;++i)
        for(j=0;j<m;++j)
            a[i][j]=tmp[i][j];
}
int main()
{
    int i,j,k;
    while(~scanf("%d%d%d%d",&n,&T,&s,&t))
    {
        prepare(dis),prepare(ans);
        for(i=0;i<mn;++i)ans[i][i]=0;
        m=0;
        memset(id,-1,sizeof(id));
        while(T--)
        {
            scanf("%d%d%d",&k,&i,&j);
            i=id[i]<0?id[i]=m++:id[i];
            j=id[j]<0?id[j]=m++:id[j];
            dis[i][j]=dis[j][i]=min(dis[i][j],(LL)k);
        }
        while(n)
        {
            if(n&1)Multi(ans,dis);
            Multi(dis,dis);
            n>>=1;
        }
        printf("%I64d
",ans[id[s]][id[t]]);
    }
    return 0;
}