Cow Relays

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.

* 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

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

 1 #include<cstdio> 2 #include<algorithm> 3 #include<map> 4 #include<cstring> 5  6 using namespace std; 7 const int MAXN = 210; 8  9 int Hash[1010];10 int n,k,q,z,tn; 11 12 int read()13 {14     int x = 0,f = 1;char ch = getchar();15     while (ch<‘0‘||ch>‘9‘) {if (ch=‘-‘) f=-1;ch = getchar(); }16     while (ch>=‘0‘&&ch<=‘9‘){x = x*10+ch-‘0‘;ch = getchar(); }17     return x*f;18 }19 20 struct Matrix{21     int a[MAXN][MAXN];22     Matrix operator * (const Matrix &x) const 23     {24         Matrix c;25         memset(c.a,0x3f,sizeof(c.a));26         for (int k=1; k<=tn; k++)27             for (int i=1; i<=tn; ++i)28                 for (int j=1; j<=tn; ++j)29                     c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]);30         return c;31      } 32 }s,ans;33 void ksm()34 { 35     ans = s;36     k--;37     for (; k; k>>=1)38     {39         if (k&1) ans = ans*s;40         s = s*s;41     }42 }43 int main()44 {45     k = read();n = read();q = read();z = read();46     memset(s.a,0x3f,sizeof(s.a));47     for (int x,y,w,i=1; i<=n; ++i)48     {49         w = read();x = read();y = read();50         if (!Hash[x]) Hash[x] = ++tn;51         if (!Hash[y]) Hash[y] = ++tn;52         s.a[Hash[x]][Hash[y]] = s.a[Hash[y]][Hash[x]] = w;53     }54     ksm();55     printf("%d",ans.a[Hash[q]][Hash[z]]);56     return 0;57 }