POJ 3368 RMQ

此题用线段树也能做 用RMQ也能做

当然RMQ的速度比线段树要快很多 

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 111111#define MAXM 11111#define INF 1000000000using namespace std;int mx[MAXN][18], Log[MAXN];struct PP{    int left, right, num;} p[MAXN];int v[MAXN], h[MAXN];int n, m, len;void rmqinit(){    for(int i = 1; i <= n; i++) mx[i][0] = p[i].num;    int m = Log[n];    for(int i = 1; i <= m; i++)        for(int j = 1; j <= n; j++)        {            mx[j][i] = mx[j][i - 1];            if(j + (1 << (i - 1)) <= n) mx[j][i] = max(mx[j][i], mx[j + (1 << (i - 1))][i - 1]);        }}int rmqmax(int l, int r){    int m = Log[r - l + 1];    return max(mx[l][m] , mx[r - (1 << m) + 1][m]);}void get(){    len = 1;    p[1].left = 1;    p[1].num = 1;    h[1] = 1;    for(int i = 2; i <= n; i++)    {        if(v[i] == v[i - 1])            p[len].num++;        else        {            p[len++].right = i - 1;            p[len].left = i;            p[len].num = 1;        }        h[i] = len;    }    p[len].right = n;}int in(){    int flag = 1;    char ch;    int a = 0;    while((ch = getchar()) == ' ' || ch == '\n');    if(ch == '-') flag = -1;    else        a += ch - '0';    while((ch = getchar()) != ' ' && ch != '\n')    {        a *= 10;        a += ch - '0';    }    return flag * a;}void out(int a){    if(a < 0)    {        putchar('-');        a = -a;    }    if(a >= 10)out(a / 10);    putchar(a % 10 + '0');}int main(){    Log[1] = 0;    for(int i = 2; i < MAXN; i++) Log[i] = Log[i >> 1] + 1;    while(scanf("%d", &n) != EOF && n)    {        scanf("%d", &m);        for(int i = 1; i <= n; i++) v[i] = in();        get();        n = len;        rmqinit();        for(int i = 1; i <= m; i++)        {            int x, y;            x= in();            y = in();            int u = h[x];            int v = h[y];            if(u == v) out(y - x + 1);            else if(u == v - 1) out(max(p[u].right - x + 1, y - p[v].left + 1));            else            {                int ans = max(p[u].right - x + 1, y - p[v].left + 1);                int t = rmqmax(u + 1, v - 1);                ans = max(ans, t);                out(ans);            }            putchar('\n');        }    }    return 0;}