无聊题目大纪实(HDU 4288 Coder)

  这神题,暴力数组可过,vector可过,线段树可过。。。时限拉的太长了,所以就成水题了。。。不过比赛的时候真的没敢用数组暴力写。。。

说说线段树的思路吧。

对树的每个节点 L, R, cnt, sum[5],分别表示左区间,右区间,这段区间上的点数,这段区间上sum[x%5]的和。

先把所有的数据都读入,然后离散化,去掉重复的数据建树。

再处理一边就可以了。

ps:叶节点只有sum[0]需要更新,sum[1...4]更新不到。。这个注意理解。。

View Code
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   x < y ? x : y
#define Max(x, y)   x < y ? y : x
#define E(x)    (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()    freopen("data.in", "r", stdin)
#define Write()   freopen("data.out", "w", stdout);

const double eps = 1e-8;
typedef long long LL;
const int inf = ~0u>>2;

using namespace std;

const int N = 100010;

struct node {
    int l;
    int r;
    int cnt;
    LL sum[5];
}tree[N<<2];

char op[N][5];
int num[N];
int t[N];
int tt[N];
int cnt, ts, p;

void build(int t, int l, int r) {
    tree[t].l = l;
    tree[t].r = r;
    tree[t].cnt = 0;
    CL(tree[t].sum, 0);
    if(l == r)  return ;
    int mid = MID(l, r);
    build(L(t), l, mid);
    build(R(t), mid + 1, r);
}

void Push_Up(int t) {
    for(int i = 0; i < 5; ++i) {
        tree[t].sum[i] = tree[L(t)].sum[i] + tree[R(t)].sum[((i - tree[L(t)].cnt)%5 + 5)%5];
    }
}

void updata(int t, int pos, int val, bool flag) {
    if(flag)    tree[t].cnt += 1;
    else    tree[t].cnt -= 1;
    if(tree[t].l == tree[t].r) {
        tree[t].sum[0] = flag ? val : 0;
        return ;
    }
    int mid = MID(tree[t].l, tree[t].r);
    if(pos <= mid)  updata(L(t), pos, val, flag);
    else    updata(R(t), pos, val, flag);
    Push_Up(t);
}

int find(int x) {
    int l = 0, r = ts - 1, mid;
    while(l <= r) {
        mid = MID(l, r);
        if(tt[mid] == x)    return mid;
        else if(tt[mid] > x)    r = mid - 1;
        else    l = mid + 1;
    }
    return -1;
}

int main() {
   // Read();

    int n, i;
    while(~scanf("%d", &n)) {
        cnt = 0;
        for(i = 0; i < n; ++i) {
            scanf("%s", op[i]);
            if(op[i][0] != 's') {
                scanf("%d", &num[i]);
                t[cnt++] = num[i];
            }
        }
        sort(t, t + cnt);
        ts = 0;
        tt[ts++] = t[0];
        for(i = 1; i < cnt; ++i) {
            if(t[i] != t[i-1])  tt[ts++] = t[i];
        }
        build(1, 0, ts - 1);
        for(i = 0; i < n; ++i) {
            if(op[i][0] == 's') {
                cout << tree[1].sum[2] << endl;
            } else if(op[i][0] == 'a') {
                p = find(num[i]);
                updata(1, p, num[i], 1);
            } else {
                p = find(num[i]);
                updata(1, p, num[i], 0);
            }
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/vongang/p/2691458.html