HDU6331Problem M. Walking Plan

传送门

分块floyd

$f[i][j][k]$表示从i走k步到j的最短路

$g[i][j][k]$表示从i走k*100步到j的最短路

$h[i][j][k]$表示从i至少走k步到j的最短路

询问从x到y至少走z步的最短路时，因为至多多走n步，用上面预处理的f,g,h可以得出答案。

 //Achen
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<vector>
 #include<cstdio>
 #include<queue>
 #include<cmath>
 #include<set>
 #include<map>
 #define Formylove return 0
 #define For(i,a,b) for(int i=(a);i<=(b);i++)
 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
 const int N=;
 typedef long long LL;
 typedef double db;
 using namespace std;
 int T,n,m,f[N][N][],g[N][N][],h[N][N][];

 template<typename T>void read(T &x)  {
     char ch=getchar(); x=; T f=;
     while(ch!='-'&&(ch<''||ch>'')) ch=getchar();
     if(ch=='-') f=-,ch=getchar();
     for(;ch>=''&&ch<='';ch=getchar()) x=x*+ch-''; x*=f;
 }

 //Walking Plan hdu6331
 int main() {
 #ifdef ANS
     freopen(".in","r",stdin);
     freopen(".out","w",stdout);
 #endif
     read(T);
     while(T--) {
         read(n); read(m);

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

         int inf=f[][][];

         For(i,,n) f[i][i][]=g[i][i][]=h[i][i][]=;

         For(i,,m) {
             int u,v,w;
             read(u); read(v); read(w);
             f[u][v][]=min(f[u][v][],w);
         }

         For(k,,) {
             For(l,,n) For(i,,n) For(j,,n)
                 f[i][j][k]=min(f[i][j][k],f[i][l][k-]+f[l][j][]);
         }

         For(i,,n) For(j,,n) g[i][j][]=f[i][j][];
         For(k,,)
             For(l,,n) For(i,,n) For(j,,n)
                 g[i][j][k]=min(g[i][j][k],g[i][l][k-]+g[l][j][]);
         For(i,,n) For(j,,n) For(k,,)
             For(l,k,) h[i][j][k]=min(h[i][j][k],f[i][j][l]);

         int q;
         read(q);
         while(q--) {
             int i,j,z;
             read(i); read(j); read(z);
             int ans=inf;
             For(k,,n) {
                 int a=z/,b=z%;
                 if(b==) a--,b=;
                 ans=min(ans,g[i][k][a]+h[k][j][b]);
             }
             if(ans>=inf) ans=-;
             printf("%d\n",ans);
         }
     }
     Formylove;
 }
 /*
 Sample Input
 2
 3 3
 1 2 1
 2 3 10
 3 1 100
 3
 1 1 1
 1 2 1
 1 3 1
 2 1
 1 2 1
 1
 2 1 1
 Sample Output
 111
 1
 11
 -1
 */