【hdu 4970】 Killing Monsters 【2014 Multi-University Training Contest 9 】

Problem Description

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

521 3 15 5 251 33 15 27 39 10

 

Sample Output

3
Hint
In the sample, three monsters with origin HP 5, 7 and 9 will survive.
 

 

这道题题意是你在一条长为N的路上有M个炮塔，每个炮塔可以对[L,R]范围内的怪物造成D点伤害，现在有K个怪物，每个怪物的生命值为H，从X的位置向N的位置走，问有多少个可以到达N。

这道题可以用树状数组或者线段树预处理出每个位置的伤害，再反向求一遍和数组，之后对每一个怪物，判断它的生命值是否大于和数组中当前位置的伤害就可以了，下面是程序：

#include<stdio.h>#include<string.h>#include<iostream>#define ll long longusing namespace std;const int N=100005;ll c[N],sum[N],l,n,m;inline ll read(){ll s=0;char c=getchar();while(c<'0'||c>'9'){c=getchar();}while(c>='0'&&c<='9'){s*=10;s+=c-'0';c=getchar();}return s;}void out(ll x){if(x>9){out(x/10);}putchar(x%10+'0');}int lowbit(int x){return x&(-x);}void add(int x,ll w){while(x<=l){c[x]+=w;x+=lowbit(x);}}ll ask(int x){ll s=0;while(x>0){s+=c[x];x-=lowbit(x);}return s;}int main(){ll i,j,w,s;while(l=read()){s=0;memset(c,0,sizeof(c));memset(sum,0,sizeof(sum));n=read();while(n--){i=read();j=read();w=read();add(i,w);add(j+1,-w);}for(i=1;i<=l;i++){sum[i]=ask(i);}for(i=l;i>=1;i--){sum[i]+=sum[i+1];}m=read();while(m--){w=read();i=read();s+=(w>sum[i]);}out(s);putchar('\n');}return 0;}