【线段树】 HDOJ 4973 A simple simulation problem.

1-n建一颗线段树,然后保存区间元素个数和区间元素最大值。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 50005
#define maxm 40005
#define eps 1e-10
#define mod 998244353
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
// head

LL maxv[maxn<<2], sum[maxn<<2];
LL mark[maxn<<2];
char ss[maxn];
int n, m;
void pushup(int o)
{
    sum[o] = sum[ls] + sum[rs];
    maxv[o] = max(maxv[ls], maxv[rs]);
}
void pushdown(int o)
{
    if(mark[o]) {
        LL tmp = powmod(2, mark[o]);
        sum[ls] *= tmp;
        sum[rs] *= tmp;
        maxv[ls] *= tmp;
        maxv[rs] *= tmp;
        mark[ls] += mark[o];
        mark[rs] += mark[o];
        mark[o] = 0;
    }
}
void build(int o, int L, int R)
{
    if(L == R) {
        sum[o] = maxv[o] = 1;
        mark[o] = 0;
        return;
    }
    mark[o] = 0;
    int mid = (L+R)>>1;
    build(lson);
    build(rson);
    pushup(o);
}
void updata(int o, int L, int R, LL ql, LL qr)
{
    if(L == R) {
        sum[o] += qr - ql + 1;
        maxv[o] = sum[o];
        return;
    }
    if(sum[o] == qr - ql + 1) {
        sum[o] *= 2;
        maxv[o] *= 2;
        mark[o]++;
        return;
    }
    pushdown(o);
    int mid = (L+R)>>1;
    LL tmp = sum[ls];
    if(tmp >= ql) updata(lson, ql, min(qr, tmp));
    if(tmp < qr) updata(rson, max(tmp+1, ql)-tmp, qr-tmp);
    pushup(o);
}
LL query(int o, int L, int R, LL ql, LL qr)
{
    if(L == R) return qr - ql + 1;
    if(sum[o] == qr - ql + 1) return maxv[o];
    pushdown(o);
    int mid = (L+R)>>1;
    LL ans = 0;
    if(sum[ls] >= ql) ans = max(ans, query(lson, ql, min(qr, sum[ls])));
    if(sum[ls] < qr) ans = max(ans, query(rson, max(sum[ls]+1, ql)-sum[ls], qr-sum[ls]));
    pushup(o);
    return ans;
}
void work(void)
{
    LL ql, qr;
    while(m--) {
        scanf("%s", ss);
        scanf("%I64d%I64d", &ql, &qr);
        if(ss[0] == 'D') updata(1, 1, n, ql, qr);
        else printf("%I64d\n", query(1, 1, n, ql, qr));
    }
}
int main(void)
{
    int _, __;
    while(scanf("%d", &_)!=EOF) {
        __ = 0;
        while(_--) {
            scanf("%d%d", &n, &m);
            build(1, 1, n);
            printf("Case #%d:\n", ++__);
            work();
        }
    }
    return 0;
}


你可能感兴趣的:(HDU)