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HDU4288-Coder(线段树+离线+离散化)

2024-07-20 23:21:34   222人阅读

Coder

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3183    Accepted Submission(s): 1254


Problem Description
　　In mathematics and computer science, an algorithm describes a set of procedures or instructions that define a procedure. The term has become increasing popular since the advent of cheap and reliable computers. Many companies now employ a single coder to write an algorithm that will replace many other employees. An added benefit to the employer is that the coder will also become redundant once their work is done. 1
　　You are now the signle coder, and have been assigned a new task writing code, since your boss would like to replace many other employees (and you when you become redundant once your task is complete).
Your code should be able to complete a task to replace these employees who do nothing all day but eating: make the digest sum.
　　By saying “digest sum” we study some properties of data. For the sake of simplicity, our data is a set of integers. Your code should give response to following operations:
　　1. add x – add the element x to the set;
　　2. del x – remove the element x from the set;
　　3. sum – find the digest sum of the set. The digest sum should be understood by

　　where the set S is written as {a1, a2, ... , ak} satisfying a1 < a2 < a3 < ... < ak 
　　Can you complete this task (and be then fired)?
------------------------------------------------------------------------------
1 See http://uncyclopedia.wikia.com/wiki/Algorithm
 

Input
　　There’re several test cases.
　　In each test case, the first line contains one integer N ( 1 <= N <= 105 ), the number of operations to process.
　　Then following is n lines, each one containing one of three operations: “add x” or “del x” or “sum”.
　　You may assume that 1 <= x <= 109.
　　Please see the sample for detailed format.
　　For any “add x” it is guaranteed that x is not currently in the set just before this operation.
　　For any “del x” it is guaranteed that x must currently be in the set just before this operation.
　　Please process until EOF (End Of File).
 

Output
　　For each operation “sum” please print one line containing exactly one integer denoting the digest sum of the current set. Print 0 if the set is empty.
 

Sample Input
9
add 1
add 2
add 3
add 4
add 5
sum
add 6
del 3
sum
6
add 1
add 3
add 5
add 7
add 9
sum
 

Sample Output
3
4
5

Hint
C++ maybe run faster than G++ in this problem.
 

Source
2012 ACM/ICPC Asia Regional Chengdu Online
 
题意： N组操作（N<=1e5），有三种操作，加一个数，删一个数，求和，其中数小于1e9，求和的是其中a是排过续的。
思路：开5棵线段树，分别代表5个余数，将N个数离散化，区间合并的时候，减掉一个偏移量即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;

int m,n;
LL sum[5][maxn<<2];
int cnt[maxn<<2];
vector<int> vi;
int demand[maxn];
int num[maxn];

void init() {
    vi.clear();
    memset(sum,0,sizeof sum);
    memset(cnt,0,sizeof cnt);
}
void input() {
    for(int i = 0; i < m; i++) {
        char st[20];
        scanf("%s",st);
        if(st[0]=='a') {
            scanf("%d",&num[i]);
            vi.push_back(num[i]);
            demand[i] = 1;
        }
        else if(st[0]=='d') {
            scanf("%d",&num[i]);
            demand[i] = 0;
        }else {
            demand[i] = 2;
        }
    }
    sort(vi.begin(),vi.end());
    n = vi.size();
}
void pushUP(int rt) {
    cnt[rt] = cnt[rt<<1]+cnt[rt<<1|1];
    int dir = cnt[rt<<1]%5;
    for(int i = 0; i < 5; i++) {
        sum[i][rt] = sum[i][rt<<1]+sum[((i-dir)%5+5)%5][rt<<1|1];
    }
}
void update(int pos,int flag,int l,int r,int rt) {
    if(l==r) {
        if(flag) {
            sum[1][rt] = vi[pos];
            cnt[rt] = 1;
        }else {
            sum[1][rt] = 0;
            cnt[rt] = 0;
        }
        return;
    }
    int mid = (l+r)>>1;
    if(pos <= mid) {
        update(pos,flag,l,mid,rt<<1);
    }else {
        update(pos,flag,mid+1,r,rt<<1|1);
    }
    pushUP(rt);
}

void solve() {

    for(int i = 0; i < m; i++) {
        if(demand[i]==2) {
            printf("%I64d\n",sum[3][1]);
        }
        else {
            int key = lower_bound(vi.begin(),vi.end(),num[i])-vi.begin();
            update(key,demand[i],0,n-1,1);
        }
    }
}
int main(){
    while(~scanf("%d",&m)) {
        init();
        input();
        solve();
    }
    return 0;
}


HDU4288-Coder(线段树+离线+离散化)


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