[USACO 07NOV]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.

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

Sample Output

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。

 #include<set>

 #include<map>

 #include<stack>

 #include<ctime>

 #include<cmath>

 #include<queue>

 #include<string>

 #include<cstdio>

 #include<vector>

 #include<cstdlib>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

 #define LL long long

 #define RE register

 #define IL inline

 using namespace std;

 const int INF=1e9;

 IL int Min(int a,int b){return a<b ? a:b;}

 int num[],pos;

 int f[][];

 int k,m,s,e,u,v,l;

 struct mat

 {

     int a[][];

     mat() {for (int i=;i<=pos;i++)for (int j=;j<=pos;j++) a[i][j]=INF;}

     mat operator * (const mat &b)

     {

         mat ans;

         for (RE int i=;i<=pos;i++)

             for (RE int j=;j<=pos;j++)

                 for (RE int k=;k<=pos;k++)

                     ans.a[i][j]=Min(ans.a[i][j],a[i][k]+b.a[k][j]);

         return ans;

     }

 };

 int main()

 {

     scanf("%d%d%d%d",&k,&m,&s,&e);

     for (RE int i=;i<=m;i++)

     {

         scanf("%d%d%d",&l,&u,&v);

         if (!num[u]) num[u]=++pos;

         if (!num[v]) num[v]=++pos;

         f[num[u]][num[v]]=f[num[v]][num[u]]=l;

     }

     mat S,T;

     for (RE int i=;i<=pos;i++) for (RE int j=;j<=pos;j++) if (f[i][j]) S.a[i][j]=T.a[i][j]=f[i][j];

     k--;

     while (k)

     {

         if (k&) S=S*T;

         k>>=;

         T=T*T;

     }

     printf("%d\n",S.a[num[s]][num[e]]);

     return ;

 }

矩乘

解法二：

利用倍增的思想。令$f[i][j][t]$表示从$i$到$j$经过$2^t$条边的最优值，做一遍$floyd$再统计答案即可。

 #include<cmath>

 #include<queue>

 #include<ctime>

 #include<stack>

 #include<cstdio>

 #include<string>

 #include<cstdlib>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

  using namespace std;

 const int INF=1e9;

 int num[],pos;

 int f[][][];

 int towards[][];

 bool t;

 int k,m,s,e,u,v,l;

 int main()

 {

     memset(f,/,sizeof(f));

     memset(towards,/,sizeof(towards));

     scanf("%d%d%d%d",&k,&m,&s,&e);

     for (int i=;i<=m;i++)

     {

         scanf("%d%d%d",&l,&u,&v);

         if (!num[u]) num[u]=++pos;

         if (!num[v]) num[v]=++pos;

         f[num[u]][num[v]][]=f[num[v]][num[u]][]=l;

     }

     int lim=log2(k);

     for (int p=;p<=lim;p++)

         for (int q=;q<=pos;q++)

             for (int i=;i<=pos;i++)

                 for (int j=;j<=pos;j++)// if (i!=j&&q!=i)

                     if (f[i][j][p]>f[i][q][p-]+f[q][j][p-])

                         f[i][j][p]=f[i][q][p-]+f[q][j][p-];

     int p=;

     towards[num[s]][t]=;

     while (k!=)

     {

         if (k&)

         {

             t=!t;

             for (int i=;i<=pos;i++)

             {

                 towards[i][t]=INF;

                 for (int j=;j<=pos;j++)

                     if (towards[i][t]>towards[j][!t]+f[i][j][p])

                         towards[i][t]=towards[j][!t]+f[i][j][p];

             }

         }

         p++;

         k=k>>;

     }

     printf("%d\n",towards[num[e]][t]);

     return ;

 }

倍增