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
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.
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
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
Case 1:
0
1
0
1
Case 2:
3
2
#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;
}
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