HDU 4031 Attack

Attack

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1904    Accepted Submission(s): 560

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

 

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

解题思路：被攻击次数=总攻击次数-防御成功次数。总攻击次数用线段树或者树状数组，防御次数暴力更新。pre数组记录上次成功防御的时间点

#include<iostream>
#include<cstdio>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Max 20005
int N,Q,t;
struct
{
    int atk;
    int cover;
}tree[Max<<2];
void push_up(int rt)
{
    tree[rt].atk=tree[rt<<1].atk+tree[rt<<1|1].atk;
}
void push_down(int rt,int m)
{
    if(tree[rt].cover)
    {
        tree[rt<<1].cover+=tree[rt].cover;
        tree[rt<<1|1].cover+=tree[rt].cover;
        tree[rt<<1].atk+=(m-(m>>1))*tree[rt].cover;
        tree[rt<<1|1].atk+=(m>>1)*tree[rt].cover;
        tree[rt].cover=0;
    }
}
void update(int L,int R,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        tree[rt].cover++;
        tree[rt].atk+=r-l+1;
        return ;
    }
    push_down(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)
        update(L,R,lson);
    if(R>m)
        update(L,R,rson);
    push_up(rt);
}
int query(int pos,int l,int r,int rt)
{
    if(l==r)
        return tree[rt].atk;
    push_down(rt,r-l+1);
    int m=(l+r)>>1;
    if(pos<=m)
        return query(pos,lson);
    else
        return query(pos,rson);
}
int main()
{
    int i,T,time,a,b,atk[Max][2],pre[Max],def[Max],ncase=1;
    char op[10];
    scanf("%d",&T);
    while(T--)
    {
        time=0;
        memset(tree,0,sizeof(tree));
        scanf("%d%d%d",&N,&Q,&t);
        memset(atk,0,sizeof(atk));
        memset(def,0,sizeof(def));
        for(i=1;i<=N;i++)
            pre[i]=1;
        printf("Case %d:\n",ncase++);
        while(Q--)
        {
            scanf("%s",op);
            if(op[0]=='A')
            {
                scanf("%d%d",&a,&b);
                atk[++time][0]=a;atk[time][1]=b;
                update(a,b,1,N,1);
            }
            else
            {
                scanf("%d",&a);
                if(t==0)
                {
                    printf("0\n");
                    continue;
                }
                for(i=pre[a];i<=time;i++)
                {
                    if(atk[i][0]<=a&&atk[i][1]>=a)
                    {
                        def[a]++;
                        pre[a]=i+t;
                        i+=t-1;
                    }
                }
                printf("%d\n",query(a,1,N,1)-def[a]);
            }
        }
    }
    return 0;
}