poj 3613 Cow Relays（矩阵连乘在图论里的应用）

Cow Relays

Time Limit: 1000MS Memory Limit: 65536KTotal 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 ≤ I_1i ≤ 1,000; 1 ≤ I_2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the length_i of each trail (1 ≤ length_i  ≤ 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: length_i , I_1i , and I_2i

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 6 411 4 64 4 88 4 96 6 82 6 93 8 9

Sample Output

10

Source

USACO 2007 November Gold

题目：http://poj.org/problem?id=3613

题意：给你一张图，要你求出边数为n的一条从s到e的最短路。。。

分析：n比较大，一开始没什么想法，后来突然想到用一个矩阵表示点之间的转换关系，然后用矩阵连乘。。。虽然跟正解有点接近但是始终没反应过来= =

其实只要把矩阵的意思换成用了几条边后，点i到j的最短距离就行。。。每次矩阵相乘都是用floyd的方式来转移的。。。

代码：

#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\n",ans[id[s]][id[t]]);    }    return 0;}