洛谷P2886 [USACO07NOV]牛继电器Cow Relays

题目描述

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.

给出一张无向连通图，求S到E经过k条边的最短路。

输入输出格式

输入格式：

* 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

输出格式：

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

输入输出样例

输入样例#1： 复制

2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9

输出样例#1： 复制

10

题解：
一句话题意：给出一个有t条边的图，求从s到e恰好经过k条边的最短路。
a:是图的邻接矩阵，f是图中任意两点直接的最短距离、

floyd算法: 
f[i][j]=min(f[i][j],f[i][k]+f[k][j]); 
一遍floyed后： f[i][j]是i到j的最短距离：中间至少1条边（连通图），最多n-1条边  

floyd算法的变形：

a:是图的邻接矩阵，a[i][j]是经过一条边的最短路径。f[i][j]k的初值为∞

f[i][j]-1=∞;

f[i][j]1=a;    //经过一条边的最短路径

f[i][j]2=min(f[i][j]2,a[i][k]+a[k][j])=a*a=a2;//经过二条边的最短路径，经过一次floyd.矩阵相乘一次,f[i][j]2初值为∞。

f[i][j]3=min(f[i][j]3,f[i][k]2+a[k][j])=a*a*a=a3;//经过三条边的最短路径，经过二次floyd。f[i][j]3初值为∞

f[i][j]4=min(f[i][j]4,f[i][k]3+a[k][j])=a4;//经过四条边的最短路径，经过三次floyd,f[i][j]4初值为∞

    ...

f[i][j]k=min(f[i][j]k,f[i][k]k-1+a[k][j])=ak;//经过k条边的最短路径，经过K-1次floyd,f[i][j]k初值为∞

而floyd的时间复杂度为O（n3）,则从的时间复杂度为O（Kn3），非常容易超时。
所以我们可以用快速幂来完成。

f[i][j]r+p=min(f[i][j]r+p,f[i][k]r+f[k][j]p)

程序：

//洛谷2886 
//（1）对角线不能设置为0，否则容易自循环。（2）数组要放在主程序外面 
//（3）K条边，不一定是最简路 
#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
using namespace std;
map<int,int>f;
const int maxn=210;
int k,t,s,e,n;
int a[maxn][maxn];
struct Matrix{
    int b[maxn][maxn];
};
Matrix A,S;
Matrix operator *(Matrix A,Matrix B){//运算符重载 
    Matrix c;
    memset(c.b,127/3,sizeof(c.b) );
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                c.b[i][j]=min(c.b[i][j],A.b[i][k]+B.b[k][j]);
    return c;
}
Matrix power(Matrix A,int k){
    if(k==0) return A;
    Matrix S=A;
    for(int i=1;i<=n;i++){
            for (int j=1;j<=n;j++)
                cout<<A.b[i][j]<<" ";
            cout<<endl;
        }
        cout<<endl;
    while(k){
        if(k&1)S=S*A;//奇数执行，偶数不执行 
        /*cout<<k<<":"<<endl;
            for(int i=1;i<=n;i++){
            for (int j=1;j<=n;j++)
                cout<<S.b[i][j]<<" ";
            cout<<endl;
        }*/
        cout<<endl;
        A=A*A;
        k=k>>1;
    
    }
    return S;
}
int main(){
    cin>>k>>t>>s>>e;
    int w,x,y;
    memset(A.b,127/3,sizeof(A.b) );// 赋初值
    n=0;
    for(int i=1;i<=t;i++){//t<100,x<1000  点不是从1开始的，可以用map离散化，给点从1开始编号。n记录共有多少个点。 
        cin>>w>>x>>y;
        if (f[x]==0)    f[x]=++n;
        if (f[y]==0)    f[y]=++n;
        A.b[f[x]][f[y]]=A.b[f[y]][f[x]]=min(w,A.b[f[y]][f[x]]);
    }
    S=power(A,k-1);//快速幂.执行k-1次floyd矩阵乘
    s=f[s];
    e=f[e];
    
    cout<<S.b[s][e];
    return 0;
}