hdu 4031 Attack 线段树

题目链接

Attack

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2330    Accepted Submission(s): 695

Problem Description

Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack American again. However, American is protected by a high wall this time, which can be treating as a segment with length N. Al Qaeda has a super weapon, every second it can attack a continuous range of the wall. American deployed N energy shield. Each one defends one unit length of the wall. However, after the shield defends one attack, it needs t seconds to cool down. If the shield defends an attack at kth second, it can’t defend any attack between (k+1)th second and (k+t-1)th second, inclusive. The shield will defend automatically when it is under attack if it is ready.

During the war, it is very important to understand the situation of both self and the enemy. So the commanders of American want to know how much time some part of the wall is successfully attacked. Successfully attacked means that the attack is not defended by the shield.

 

 

Input

The beginning of the data is an integer T (T ≤ 20), the number of test case.
The first line of each test case is three integers, N, Q, t, the length of the wall, the number of attacks and queries, and the time each shield needs to cool down.
The next Q lines each describe one attack or one query. It may be one of the following formats
1. Attack si ti
  Al Qaeda attack the wall from si to ti, inclusive. 1 ≤ si ≤ ti ≤ N
2. Query p
  How many times the pth unit have been successfully attacked. 1 ≤ p ≤ N
The kth attack happened at the kth second. Queries don’t take time.
1 ≤ N, Q ≤ 20000
1 ≤ t ≤ 50

 

 

Output

For the ith case, output one line “Case i: ” at first. Then for each query, output one line containing one integer, the number of time the pth unit was successfully attacked when asked.

 

 

Sample Input

2 3 7 2 Attack 1 2 Query 2 Attack 2 3 Query 2 Attack 1 3 Query 1 Query 3 9 7 3 Attack 5 5 Attack 4 6 Attack 3 7 Attack 2 8 Attack 1 9 Query 5 Query 3

 

 

Sample Output

Case 1: 0 1 0 1 Case 2: 3 2

 

基础的是线段树区间更新+单点查询， 因为有那个保护的时间， 所以开一个辅助的数组来记录状态， 具体的看代码。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 2e4+5;
int sum[maxn<<2], add[maxn<<2];
void pushUp(int rt) {
    sum[rt] = sum[rt<<1]+sum[rt<<1|1];
}
void pushDown(int rt, int m) {
    if(add[rt]) {
        sum[rt<<1] += (m-(m>>1))*add[rt];
        sum[rt<<1|1] += (m>>1)*add[rt];
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        add[rt] = 0;
    }
}
void update(int L, int R, int l, int r, int rt) {
    if(L<=l&&R>=r) {
        sum[rt] += r-l+1;
        add[rt]++;
        return ;
    }
    pushDown(rt, r-l+1);
    int m = l+r>>1;
    if(L<=m)
        update(L, R, lson);
    if(R>m)
        update(L, R, rson);
    pushUp(rt);
}
int query(int p, int l, int r, int rt) {
    if(l == r) {
        return sum[rt];
    }
    pushDown(rt, r-l+1);
    int m = l+r>>1;
    if(p<=m)
        return query(p, lson);
    else
        return query(p, rson);
}
int time[maxn], num[maxn];
pll att[maxn];
int main()
{
    int t, n, q, k, x, y;
    cin>>t;
    char s[10];
    for(int casee = 1; casee<=t; casee++) {
        printf("Case %d:\n", casee);
        mem(add);
        mem(num);
        mem(sum);
        mem(time);
        int cnt = 0;
        cin>>n>>q>>k;
        while(q--) {
            scanf("%s%d", s, &x);
            if(s[0] == 'A') {
                scanf("%d", &y);
                update(x, y, 1, n, 1);
                att[cnt++] = mk(x, y);
            } else {
                if(k == 1) {
                    puts("0");
                    continue;
                }
                int tmp = query(x, 1, n, 1);
                for(int i = time[x]; i<cnt; i++) {
                    if(x<=att[i].se&&x>=att[i].fi) {
                        num[x]++;
                        time[x] = i+k;
                        i += k-1;
                    }
                }
                printf("%d\n", tmp-num[x]);
            }
        }
    }
    return 0;
}