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 descriptiontoday 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, random Sive. 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.
 
Inputthe 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, random Sive. 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
 
Outputfor 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

23 7 2Attack 1 2Query 2Attack 2 3Query 2Attack 1 3Query 1Query 39 7 3Attack 5 5Attack 4 6Attack 3 7Attack 2 8Attack 1 9Query 5Query 3

 
Sample output

Case 1:0101Case 2:32

 
Sourcethe 36th ACM/ICPC Asia Regional Chengdu site -- online contest Solution: number of attacks = Total number of attacks-number of successful defenses, total number of attacks using a line tree or tree array, brute force update of the number of defenses, pre Array records the time point of the last successful defense

#include<iostream>#include<cstdio>#include<cstring>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define Max 20005int 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;}

 

HDU 4031 attack