POJ 3368 Frequent values(rmq)
离散化后rmq，新白书上对这个题有讲解。
1 #include <iostream> 2 #include <cstring> 3 #include <cmath> 4 #include <cstdio> 5 using namespace std; 6 #define N 100000 7 int p[N]; 8 int dp[22][N+1]; 9 int hash[3*N]; 10 int que[N+1]; 11 int L[N+1]; 12 int R[N+1]; 13 int o[N+1]; 14 void init_rmq(int n) 15 { 16 int i,j; 17 for(i = 1;i <= n;i ++) 18 dp[0][i] = o[i]; 19 for(i = 1;(1<<i) <= n;i ++) 20 { 21 for(j = 1;j + (1<<(i-1)) <= n;j ++) 22 { 23 dp[i][j] = max(dp[i-1][j],dp[i-1][j+(1<<(i-1))]); 24 } 25 } 26 } 27 int rmq(int x,int y) 28 { 29 if(x > y) return 0; 30 int k = log(y-x+1.0)/log(2.0); 31 return max(dp[k][x],dp[k][y-(1<<k)+1]); 32 } 33 int query(int x,int y) 34 { 35 int temp,nx,ny; 36 if(p[x] == p[y]) 37 return y-x+1; 38 nx = hash[p[x]+N]; 39 ny = hash[p[y]+N]; 40 temp = max(R[nx]-x+1,y-L[ny]+1); 41 temp = max(temp,rmq(nx+1,ny-1)); 42 return temp; 43 } 44 int main() 45 { 46 int n,m,i,temp,num,x,y; 47 while(scanf("%d",&n)!=EOF) 48 { 49 if(n == 0) break; 50 scanf("%d",&m); 51 for(i = 1;i <= n;i ++) 52 { 53 scanf("%d",&p[i]); 54 } 55 L[1] = 1; 56 num = 1; 57 temp = 1; 58 hash[p[1]+N] = 1; 59 for(i = 2;i <= n;i ++) 60 { 61 if(p[i] != p[i-1]) 62 { 63 R[num] = i-1; 64 o[num] = temp; 65 hash[p[i-1]+N] = num; 66 temp = 1; 67 num ++; 68 L[num] = i; 69 } 70 else 71 { 72 temp ++; 73 } 74 } 75 R[num] = i-1; 76 o[num] = temp; 77 hash[p[i-1]+N] = num; 78 init_rmq(num); 79 while(m--) 80 { 81 scanf("%d%d",&x,&y); 82 printf("%d\n",query(x,y)); 83 } 84 } 85 return 0; 86 }
