hdu 4031 attack 线段树区间更新

Attack

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2496    Accepted Submission(s): 788

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

 

/*
hdu 4031 attack 线段树区间更新

problem:
每个位置有一个防御塔，每次敌人会对一个区间进行攻击。防御塔有一个冷却时间t。
问某个时间时，某个位置被成功攻击的次数（没有被防御）

solve:
主要是怎么处理这个冷却值的问题，区间攻击都可以使用 线段树区间更新来解决。
最开始考虑的是通过 更新来维护每个点，那每次就要更新[1,n]，感觉应该会超时吧
然后想到是 通过每次-1来体现时间的变化，但是无法知道每个点攻击成功的情况

所以直接暴力搞了- -
通过线段树记录被攻击了多少次，然后通过记录可以知道防御成功了多少次
于是就能得出答案。

hhh-2016-08-05 21:16:55
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#include <map>
#include <queue>
#include <vector>
#include <set>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;
typedef long long ll;
const int maxn=100000 + 10;
const int INF=0x3f3f3f3f;
const int mod = 1e9+7;
int n;

struct node
{
    int l,r;
    int val;
    int len ;
    int mid()
    {
        return (l+r)>> 1;
    }
} tree[maxn<<2];

void push_up(int i)
{
}

void build(int i,int l,int r)
{
    tree[i].l = l;
    tree[i].r = r;
    tree[i].val = tree[i].len = 0;
    if(l == r)
    {
        return ;
    }
    int mid = tree[i].mid();
    build(lson,l,mid);
    build(rson,mid+1,r);
    push_up(i);
}

void push_down(int i)
{
    if(tree[i].val)
    {
        tree[lson].val += tree[i].val;
        tree[rson].val += tree[i].val;
        tree[i].val = 0;
    }
}

void Insert(int i,int l,int r,int val)
{
    if(tree[i].l >= l && tree[i].r <= r)
    {
        tree[i].val += val;
        push_up(i);
        return ;
    }
    push_down(i);
    int mid = tree[i].mid();
    if(l <= mid)
        Insert(lson,l,r,val);
    if(r > mid)
        Insert(rson,l,r,val);
    push_up(i);
}

int query(int i,int k)
{
    if(tree[i].l == tree[i].r && tree[i].l == k)
    {
        return tree[i].val;
    }
    push_down(i);
    int mid = tree[i].mid();
    if(k <= mid)
        return query(lson,k);
    else
        return query(rson,k);
}
struct Point
{
    int l,r;
    Point()
    {

    }
    Point(int a,int b)
    {
        l = a,r = b;
    }
};
Point pt[maxn];
char op[5];

int main()
{
    int T,n,m,k;
    int cas = 1;
    int a,b;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        int cnt = 0;
        build(1,1,n);
        printf("Case %d:\n",cas++);
        for(int i = 1;i <= m;i++)
        {
            scanf("%s",op);
            if(op[0] == 'Q')
            {
                int num = 0;
                scanf("%d",&a);
                for(int i = 0;i < cnt;)
                {
                    if( a>= pt[i].l && a <= pt[i].r)
                    {
                        num++;
                        i += k;
                    }
                    else
                    {
                        i++;
                    }
                }
                printf("%d\n",query(1,a) - num);
            }
            else
            {
                scanf("%d%d",&a,&b);
                Insert(1,a,b,1);
                pt[cnt++] = Point(a,b);
            }
        }
    }
    return 0;
}