洛谷P2886 [USACO07NOV]牛繼電器Cow Relays

題目描述

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.

給出一張無向連通圖，求S到E通過k條邊的最短路。

輸入輸出格式

輸入格式：

 

* 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

 

輸出格式：

 

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

 

輸入輸出樣例

輸入樣例#1：  複製

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

輸出樣例#1：  複製

10

題解：
一句話題意：給出一個有t條邊的圖，求從s到e剛好通過k條邊的最短路。
a:是圖的鄰接矩陣，f是圖中任意兩點直接的最短距離、

floyd算法: 
f[i][j]=min(f[i][j],f[i][k]+f[k][j]); 
一遍floyed後： f[i][j]是i到j的最短距離：中間至少1條邊（連通圖），最多n-1條邊 

floyd算法的變形：

a:是圖的鄰接矩陣，a[i][j]是通過一條邊的最短路徑。f[i][j]k的初值爲∞

f[i][j]-1=∞;

f[i][j]1=a;    //通過一條邊的最短路徑

f[i][j]2=min(f[i][j]2,a[i][k]+a[k][j])=a*a=a2;//通過二條邊的最短路徑，通過一次floyd.矩陣相乘一次,f[i][j]2初值爲∞。

f[i][j]3=min(f[i][j]3,f[i][k]2+a[k][j])=a*a*a=a3;//通過三條邊的最短路徑，通過二次floyd。f[i][j]3初值爲∞

f[i][j]4=min(f[i][j]4,f[i][k]3+a[k][j])=a4;//通過四條邊的最短路徑，通過三次floyd,f[i][j]4初值爲∞

    ...

f[i][j]k=min(f[i][j]k,f[i][k]k-1+a[k][j])=ak;//通過k條邊的最短路徑，通過K-1次floyd,f[i][j]k初值爲∞

而floyd的時間複雜度爲O（n3）,則從的時間複雜度爲O（Kn3），很是容易超時。
因此咱們能夠用快速冪來完成。

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）對角線不能設置爲0，不然容易自循環。（2）數組要放在主程序外面 
//（3）K條邊，不必定是最簡路 
#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){//運算符重載 
    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;//奇數執行，偶數不執行 
        /*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) );// 賦初值
    n=0;
    for(int i=1;i<=t;i++){//t<100,x<1000  點不是從1開始的，能夠用map離散化，給點從1開始編號。n記錄共有多少個點。 
        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);//快速冪.執行k-1次floyd矩陣乘
    s=f[s];
    e=f[e];
    
    cout<<S.b[s][e];
    return 0;
}