POJ3613 Cow Relays [矩阵乘法 floyd类似]

Cow Relays

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7335		Accepted: 2878

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 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

---

b在大很多也可以

求经过b条边的最短路

貌似直接想floyd不太明白了，一遍floyd怎么是经过一条边的最短路呢？

还是从矩阵乘法考虑，构造一个向量表示距离，不停乘邻接矩阵

定义一种新矩阵乘法，不是加起来而是求最小值

 

初始化时用0x3f

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=;
typedef long long ll;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}
int b,n,m,s,t,u,v,w;
int mp[];
struct Mat{
    int a[N][N];
    Mat(){memset(a,0x3f,sizeof(a));}
}G,ans;
Mat operator *(Mat A,Mat B){
    Mat C;
    for(int k=;k<=n;k++)
        for(int i=;i<=n;i++) if(A.a[i][k])
            for(int j=;j<=n;j++) if(B.a[k][j])
                C.a[i][j]=min(C.a[i][j],A.a[i][k]+B.a[k][j]);
    return C;
}
int main(){
    //freopen("in.txt","r",stdin);
    b=read();m=read();s=read();t=read();
    if(!mp[s]) mp[s]=++n;s=mp[s];
    if(!mp[t]) mp[t]=++n;t=mp[t];
    for(int i=;i<=m;i++){
        w=read();u=read();v=read();
        if(!mp[u]) mp[u]=++n;u=mp[u];
        if(!mp[v]) mp[v]=++n;v=mp[v];
        G.a[u][v]=G.a[v][u]=w;
    }
    ans=G;b--;
    for(;b;b>>=,G=G*G)
        if(b&) ans=ans*G;
    printf("%d",ans.a[s][t]);
}