poj 3613 Cow Relays（矩阵连乘在图论里的应用）

最新推荐文章于 2021-07-14 09:09:43 发布

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最新推荐文章于 2021-07-14 09:09:43 发布

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分类专栏：图论文章标签： distance structure integer each output input

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本文介绍了一种解决特定图论问题的方法——寻找一条通过指定数量的边连接两个点的最短路径。采用矩阵连乘的方式高效求解，并提供了一个完整的实现示例。

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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 ≤ 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

Sample Output

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;
}