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.
给出一张无向连通图，求S到E经过k条边的最短路。
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
题解
解法一：
考虑有用的点数很少，我们可以哈希一下，建立邻接矩阵，矩阵加速求出经过$N$条边的从$S$到$T$的最短路。
1 #include<set> 2 #include<map> 3 #include<stack> 4 #include<ctime> 5 #include<cmath> 6 #include<queue> 7 #include<string> 8 #include<cstdio> 9 #include<vector> 10 #include<cstdlib> 11 #include<cstring> 12 #include<iostream> 13 #include<algorithm> 14 #define LL long long 15 #define RE register 16 #define IL inline 17 using namespace std; 18 const int INF=1e9; 19 20 IL int Min(int a,int b){return a<b ? a:b;} 21 22 int num[1005],pos; 23 int f[205][205]; 24 int k,m,s,e,u,v,l; 25 struct mat 26 { 27 int a[205][205]; 28 mat() {for (int i=1;i<=pos;i++)for (int j=1;j<=pos;j++) a[i][j]=INF;} 29 mat operator * (const mat &b) 30 { 31 mat ans; 32 for (RE int i=1;i<=pos;i++) 33 for (RE int j=1;j<=pos;j++) 34 for (RE int k=1;k<=pos;k++) 35 ans.a[i][j]=Min(ans.a[i][j],a[i][k]+b.a[k][j]); 36 return ans; 37 } 38 }; 39 40 int main() 41 { 42 scanf("%d%d%d%d",&k,&m,&s,&e); 43 for (RE int i=1;i<=m;i++) 44 { 45 scanf("%d%d%d",&l,&u,&v); 46 if (!num[u]) num[u]=++pos; 47 if (!num[v]) num[v]=++pos; 48 f[num[u]][num[v]]=f[num[v]][num[u]]=l; 49 } 50 mat S,T; 51 for (RE int i=1;i<=pos;i++) for (RE int j=1;j<=pos;j++) if (f[i][j]) S.a[i][j]=T.a[i][j]=f[i][j]; 52 k--; 53 while (k) 54 { 55 if (k&1) S=S*T; 56 k>>=1; 57 T=T*T; 58 } 59 printf("%d\n",S.a[num[s]][num[e]]); 60 return 0; 61 }