一维模板
const int maxn=50001; class CRMQ { private: int Max[20][maxn]; int Min[20][maxn]; int idx[maxn]; public: int val[maxn]; void initRMQ(int n) { idx[0]=-1; for (int i=1; i<=n; i++){ idx[i]=(i&(i-1))?idx[i-1]:idx[i-1]+1; Min[0][i]=Max[0][i]=val[i]; } for (int i=1; i<=idx[n]; i++){ int limit=n+1-(1<<i); for (int j=1; j<=limit; j++){ Min[i][j]=min(Min[i-1][j],Min[i-1][j+(1<<i>>1)]); Max[i][j]=max(Max[i-1][j],Max[i-1][j+(1<<i>>1)]); } } } int getval(int a,int b) { int t=idx[b-a+1]; b-=(1<<t)-1; return max(Max[t][a],Max[t][b])-min(Min[t][a],Min[t][b]); } }; CRMQ sol;
模板题目:POJ3264
AC代码:
/** head-file **/ #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstring> #include <string> #include <vector> #include <queue> #include <stack> #include <list> #include <set> #include <map> #include <algorithm> /** define-for **/ #define REP(i, n) for (int i=0;i<int(n);++i) #define FOR(i, a, b) for (int i=int(a);i<int(b);++i) #define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i) #define REP_1(i, n) for (int i=1;i<=int(n);++i) #define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i) #define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i) #define REP_N(i, n) for (i=0;i<int(n);++i) #define FOR_N(i, a, b) for (i=int(a);i<int(b);++i) #define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i) #define REP_1_N(i, n) for (i=1;i<=int(n);++i) #define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i) #define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i) /** define-useful **/ #define clr(x,a) memset(x,a,sizeof(x)) #define sz(x) int(x.size()) #define see(x) cerr<<#x<<" "<<x<<endl #define se(x) cerr<<" "<<x #define pb push_back #define mp make_pair /** test **/ #define Display(A, n, m) { REP(i, n){ REP(j, m) cout << A[i][j] << " "; cout << endl; } } #define Display_1(A, n, m) { REP_1(i, n){ REP_1(j, m) cout << A[i][j] << " "; cout << endl; } } using namespace std; /** typedef **/ typedef long long LL; /** Add - On **/ const int direct4[4][2]= { {0,1},{1,0},{0,-1},{-1,0} }; const int direct8[8][2]= { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} }; const int direct3[6][3]= { {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} }; const int MOD = 1000000007; const int INF = 0x3f3f3f3f; const long long INFF = 1LL << 60; const double EPS = 1e-9; const double OO = 1e15; const double PI = acos(-1.0); //M_PI; const int maxn=50001; class CRMQ { private: int Max[20][maxn]; int Min[20][maxn]; int idx[maxn]; public: int val[maxn]; void initRMQ(int n) { idx[0]=-1; for (int i=1; i<=n; i++){ idx[i]=(i&(i-1))?idx[i-1]:idx[i-1]+1; Min[0][i]=Max[0][i]=val[i]; } for (int i=1; i<=idx[n]; i++){ int limit=n+1-(1<<i); for (int j=1; j<=limit; j++){ Min[i][j]=min(Min[i-1][j],Min[i-1][j+(1<<i>>1)]); Max[i][j]=max(Max[i-1][j],Max[i-1][j+(1<<i>>1)]); } } } int getval(int a,int b) { int t=idx[b-a+1]; b-=(1<<t)-1; return max(Max[t][a],Max[t][b])-min(Min[t][a],Min[t][b]); } }; CRMQ sol; int main() { int n,m; scanf("%d%d",&n,&m); REP_1(i,n) scanf("%d",&sol.val[i]); sol.initRMQ(n); while (m--) { int a,b; scanf("%d%d",&a,&b); printf("%d ",sol.getval(a,b)); } return 0; }--------
二维模板
模板题目:
AC代码: