POJ3613 Cow Relays

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6726		Accepted: 2626

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

 

经过n条边的最短路径。

好像可以贪心，先找到一条最短路，然后随便找个最小环绕圈圈，我没有尝试。

 

使用了标准解法，矩阵乘法+倍增floyd

floyd是枚举k点，更新i到j的最短路径，更新后实际上就是从原本的走一条路到达更新成了走两条路到达。

根据这个思想，我们可以倍增跑floyd，就能得出走2^k条边从i到j的最短路径。

 

代码如下：（a[i]里存的就是走2^i条边的最短路邻接矩阵)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int mxn=200;
int mp[2000],cnt=0;
int m;
struct edge{
    int x,y;
    int len;
}e[mxn];
struct M{
    int v[200][200];
    M(){memset(v,50,sizeof v);}
    friend M operator *(M a,M b){
        M c;
        int i,j,k;
        for(i=0;i<=m;i++)
          for(j=0;j<=m;j++)
            for(k=0;k<=m;k++){
                c.v[i][j]=min(c.v[i][j],a.v[i][k]+b.v[k][j]);
            }
        return c;
    }
}a[120];
int n,t,S,E;
int main(){
    scanf("%d%d%d%d",&n,&t,&S,&E);
    int i,j;
    memset(mp,-1,sizeof mp);
    int u,v;
    for(i=1;i<=t;i++){
        scanf("%d%d%d",&e[i].len,&u,&v);
        if(mp[u]==-1)mp[u]=++cnt;
        if(mp[v]==-1)mp[v]=++cnt;
        e[i].x=mp[u];e[i].y=mp[v];
    }
    for(i=1;i<=t;i++){
        int x=e[i].x;int y=e[i].y;
        a[0].v[x][y]=a[0].v[y][x]=e[i].len;
    }
    if(mp[S]==-1)mp[S]=++cnt;
    if(mp[E]==-1)mp[E]=++cnt;
    m=cnt;
    S=mp[S];E=mp[E];
    for(i=1;i<=20;i++){//倍增 
        a[i]=a[i-1]*a[i-1];
    }
    M ans=a[0];
    n--;
    for(i=20;i>=0;i--){
        if((1<<i)&n)ans=ans*a[i];
    }
    printf("%d\n",ans.v[S][E]);
    return 0;
}