【floyd+矩阵乘法】POJ 3613 Cow Relays

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 ≤ 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.

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: lengthi , I1i , and I2i

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

分析

题意：给定一个T(2 <= T <= 100)条边的无向图，求S到E恰好经过N(2 <= N <= 1000000)条边的最短路。
分析：大致思路就是floyd+矩阵乘法。我们令C[S][E]表示S点到E点正好经过N条边的路径数。
接下来用Floyd每次使用一个中间点k去更新S,E之间的距离，那么更新成功表示S,E之间恰有一个点k时的最短路。我们做n次这样的操作就能够得出结果了。
我们用c[S][E]=max(c[S][E],a[S][k]+a[k][E])来进行路径长度更新。第二次将c[S][E]拷贝回到a[S][E]当中，并将c[S][E]重新置为inf，再做一次，则是在原来的基础上在S,E之间再用一个点k来松弛，这时候S,E之间实际上已经是两个点了，之后重复这么做就好了.

代码

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

const int inf = 0x7f7f7f7f;

const int maxn = ;

int K,M,S,T;

int v[maxn],cnt,map[maxn][maxn],used[maxn];

int ans[maxn][maxn],dis[maxn][maxn],tmp[maxn][maxn];

void floyd(int c[][maxn],int a[][maxn],int b[maxn][maxn]){

       int i,j,k;

       for(k=;k<cnt;k++){

              for(i=;i<cnt;i++){

                     for(j=;j<cnt;j++){

                            if(c[v[i]][v[j]]>a[v[i]][v[k]]+b[v[k]][v[j]])

                                   c[v[i]][v[j]]=a[v[i]][v[k]]+b[v[k]][v[j]];

                     }

              }

       }

}

void copy(int a[][maxn],int b[][maxn]){

       int i,j;

       for(i=;i<cnt;i++){

              for(j=;j<cnt;j++){

                     a[v[i]][v[j]]=b[v[i]][v[j]];

                     b[v[i]][v[j]]=inf;

              }

       }

}

void solve(int k){

       while(k){

              if(k%){

                     floyd(dis,ans,map);

                     copy(ans,dis);

              }

              floyd(tmp,map,map);

              copy(map,tmp);

              k=k/;

       }

}

int main(){

       int i,j;

       int x,y,val;

       while(scanf("%d%d%d%d",&K,&M,&S,&T)==){

                for(i=;i<=;i++){

                    for(j=;j<=;j++){

                        map[i][j]=inf;

                        ans[i][j]=inf;

                        tmp[i][j]=inf;

                        dis[i][j]=inf;

                    }

                    ans[i][i]=;

                }

                memset(used,,sizeof(used));

                cnt=;

                for(i=;i<M;i++){

                    scanf("%d%d%d",&val,&x,&y);

                    if(map[x][y]>val){

                            map[x][y]=val;

                            map[y][x]=map[x][y];

                    }

                    if(!used[x]){

                            used[x]=;

                            v[cnt++]=x;

                    }

                    if(!used[y]){

                            used[y]=;

                            v[cnt++]=y;

                    }

            }

            solve(K);

            printf("%d\n",ans[S][T]);

       }

       return ;

}

巴特西