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 ≤ 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
10
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[205][205]; Martix operator * (Martix &rhs){ Martix c; memset(c.a,0x3f,sizeof(c.a)); for(int i=1;i<=tot;++i) for(int j=1;j<=tot;++j) for(int k=1;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=0; 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&1) ans=ans*base,--k; k>>=1; base=base*base; } printf("%d\n",ans.a[mp[st]][mp[ed]]); }