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。

  1 //Achen
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<cstdlib>
  6 #include<vector>
  7 #include<cstdio>
  8 #include<queue>
  9 #include<cmath>
 10 #include<set>
 11 #include<map>
 12 #define Formylove return 0
 13 #define For(i,a,b) for(int i=(a);i<=(b);i++)
 14 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
 15 const int N=55;
 16 typedef long long LL;
 17 typedef double db;
 18 using namespace std;
 19 int T,n,m,f[N][N][205],g[N][N][205],h[N][N][205];
 20 
 21 template<typename T>void read(T &x)  {
 22     char ch=getchar(); x=0; T f=1;
 23     while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
 24     if(ch=='-') f=-1,ch=getchar();
 25     for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
 26 }
 27 
 28 //Walking Plan hdu6331
 29 int main() {
 30 #ifdef ANS
 31     freopen(".in","r",stdin);
 32     freopen(".out","w",stdout);
 33 #endif
 34     read(T);
 35     while(T--) {
 36         read(n); read(m);
 37         
 38         memset(f,127/3,sizeof(f));
 39         memset(g,127/3,sizeof(g));
 40         memset(h,127/3,sizeof(h));
 41         
 42         int inf=f[0][0][0];
 43         
 44         For(i,1,n) f[i][i][0]=g[i][i][0]=h[i][i][0]=0;
 45         
 46         For(i,1,m) {
 47             int u,v,w;
 48             read(u); read(v); read(w);
 49             f[u][v][1]=min(f[u][v][1],w);
 50         }
 51         
 52         For(k,2,200) {
 53             For(l,1,n) For(i,1,n) For(j,1,n) 
 54                 f[i][j][k]=min(f[i][j][k],f[i][l][k-1]+f[l][j][1]);    
 55         }
 56         
 57         For(i,1,n) For(j,1,n) g[i][j][1]=f[i][j][100];
 58         For(k,2,100) 
 59             For(l,1,n) For(i,1,n) For(j,1,n)
 60                 g[i][j][k]=min(g[i][j][k],g[i][l][k-1]+g[l][j][1]);
 61         For(i,1,n) For(j,1,n) For(k,1,100) 
 62             For(l,k,200) h[i][j][k]=min(h[i][j][k],f[i][j][l]);
 63         
 64         int q;
 65         read(q);
 66         while(q--) {
 67             int i,j,z;
 68             read(i); read(j); read(z);
 69             int ans=inf;
 70             For(k,1,n) {
 71                 int a=z/100,b=z%100;
 72                 if(b==0) a--,b=100;
 73                 ans=min(ans,g[i][k][a]+h[k][j][b]);
 74             }
 75             if(ans>=inf) ans=-1;
 76             printf("%d
",ans);
 77         }
 78     }
 79     Formylove;
 80 }
 81 /*
 82 Sample Input
 83 2        
 84 3 3    
 85 1 2 1
 86 2 3 10
 87 3 1 100
 88 3        
 89 1 1 1
 90 1 2 1
 91 1 3 1
 92 2 1    
 93 1 2 1
 94 1        
 95 2 1 1
 96 Sample Output
 97 111
 98 1
 99 11
100 -1
101 */
View Code
原文地址:https://www.cnblogs.com/Achenchen/p/9550356.html