Kingdom Rush is a popular TD game, in which you should build some towers to protect your kingdom from monsters. And now another wave of monsters is coming and you need again to know whether you can get through it.
The path of monsters is a straight line, and there are N blocks on it (numbered from 1 to N continuously). Before enemies come, you have M towers built. Each tower has an attack range [L, R], meaning that it can attack all enemies in every block i, where L<=i<=R. Once a monster steps into block i, every tower whose attack range include block i will attack the monster once and only once. For example, a tower with attack range [1, 3] will attack a monster three times if the monster is alive, one in block 1, another in block 2 and the last in block 3.
A witch helps your enemies and makes every monster has its own place of appearance (the ith monster appears at block Xi). All monsters go straightly to block N.
Now that you know each monster has HP Hi and each tower has a value of attack Di, one attack will cause Di damage (decrease HP by Di). If the HP of a monster is decreased to 0 or below 0, it will die and disappear.
Your task is to calculate the number of monsters surviving from your towers so as to make a plan B.
Input
The input contains multiple test cases.
The first line of each case is an integer N (0 < N <= 100000), the number of blocks in the path. The second line is an integer M (0 < M <= 100000), the number of towers you have. The next M lines each contain three numbers, Li, Ri, Di (1 <= Li <= Ri <= N, 0 < Di <= 1000), indicating the attack range [L, R] and the value of attack D of the ith tower. The next line is an integer K (0 < K <= 100000), the number of coming monsters. The following K lines each contain two integers Hi and Xi (0 < Hi <= 10^18, 1 <= Xi <= N) indicating the ith monster’s live point and the number of the block where the ith monster appears.
The input is terminated by N = 0.
Output
Output one line containing the number of surviving monsters.
Sample Input
5
2
1 3 1
5 5 2
5
1 3
3 1
5 2
7 3
9 1
0
Sample Output
3
Hint
In the sample, three monsters with origin HP 5, 7 and 9 will survive.
题意：有n个格子线性排列，有m个攻击塔。每个塔可在L到R范围内的每个格子里造成Z点伤害。有K个怪物，每只怪物在POS处出生，并有HP的血量，问最后能活着走出N个格子的怪物有多少。
格子数为树状数组大小，数组内存储每个格子造成的伤害，对于区间伤害不能单点更新，会超时，只能区间更新，在区间左边界L更新Z，右边界R+1更新-Z
因此当sum函数求和时，从1到x的计数，其他区间的伤害将被边界的-Z抵消，而到达X时因为还未到达边界-Z抵消，所以将得到此处格子的伤害，即使重叠覆盖的区间也能得到伤害
attact数组表示从格子pos处到达终点的总伤害，计算时从n倒着求到1，对于 每个POS的总伤害，由i-1加sum(i)递推而来
最终输入每个怪物的HP和出生点时，只需比较该怪物的血量是否能承受从出生点到终点的伤害总和，即可判断该怪物是否能存活
代码如下：
#include<stdio.h>
#include<string.h>
#define LL long long
LL tree[100005],attact[100005];
int lowbit(int x)
{
return x&(-x);
}
void update(int p,int x)///更新时从x到n
{
while(p<=n)
{
tree[p]+=x;
p+=lowbit(p);
}
}
LL sum(int x)///求和时从X到1
{
LL sum=0;
while(x>0)
{
sum+=tree[x];
x-=lowbit(x);
}
return sum;
}
int main()
{
int l,r,z;
while(scanf("%d",&n)!=EOF)
{
if(n==0)break;
scanf("%d",&m);
memset(tree,0,sizeof(tree));
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&l,&r,&z);
update(l,z);
update(r+1,-z);
}
attact[n+1]=0;
for(int i=n;i>=1;i--) attact[i]=attact[i+1]+sum(i);
int k,pos,ans=0;
LL hp;
scanf("%d",&k);
while(k--)
{
scanf("%lld%d",&hp,&pos);
if(hp>attact[pos])ans++;
}
printf("%d
",ans);
}
}