poj3613 求经过n条边的最短路 ----矩阵玩出新高度。

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

Sample Output

Source

http://blog.csdn.net/monster__yi/article/details/51069236 感谢题解

矩阵mx[i][j]表示已经有一条i->j的边，然后在和基础矩阵进行运算，那么mx[j][k]，就代表再走一条边从i到j，满足了每次只走一条边的条件

#include<cstdio>

#include<cstring>

#include<map>

using namespace std;

map<int,int>mp;

int tot,k;

struct Martix{

    int a[][];

    Martix operator * (Martix &rhs){

    Martix c;

    memset(c.a,0x3f,sizeof(c.a));

    for(int i=;i<=tot;++i)

        for(int j=;j<=tot;++j)

        for(int k=;k<=tot;++k)

        c.a[i][j]=min(c.a[i][j],a[i][k]+rhs.a[k][j]);

    return c;

    }

}base,ans;

int main(){

     int u,v,x,m,st,ed;

     tot=;

     memset(base.a,0x3f,sizeof(base.a));

     scanf("%d%d%d%d",&k,&m,&st,&ed);

     while(m--){

        scanf("%d%d%d",&x,&u,&v);

        if(!mp[u]) mp[u]=++tot;

        if(!mp[v]) mp[v]=++tot;

        base.a[mp[u]][mp[v]]=base.a[mp[v]][mp[u]]=x;

     }

     ans=base;

     --k;

     while(k){

        if(k&) ans=ans*base,--k;

        k>>=;

        base=base*base;

     }

     printf("%d\n",ans.a[mp[st]][mp[ed]]);

}

巴特西

poj3613 求经过n条边的最短路 ----矩阵玩出新高度。

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巴特西

poj3613 求经过n条边的最短路 ----矩阵玩出新高度 。

最新文章

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poj3613 求经过n条边的最短路 ----矩阵玩出新高度。