2014多校1011--hdu--4097--Killing Monsters（塔防，线段树超时。。）

Killing Monsters

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 107    Accepted Submission(s): 54

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

   
   
   
   

    
    
    
    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. 
    
    
    
    
     
   
   
   
   

 塔防问题，给出长度n的路径，m个塔，每个塔有攻击范围和伤害值，给出m个怪的出现血量和地点。问有几只怪可以走完全程

不能使用线段树，树状数组维护，10000*log（10000）*多组输入一定会超时

定义数组c，如果有i塔的攻击范围是[l,r]伤害是d，那么在c[l]处+d,在c[r+1]处-d，这样由1累积到n，当累加到k点的值是就是在k点怪一共可以收到的伤害，再从n到1累加回去，使得每个点存储了该点到n一共会受到的伤害，只要血量大于伤害就可以走完，统计数目

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 110000
#define LL __int64
#define lmin 1
#define rmax n
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define root lmin,rmax,1
#define now l,r,rt
#define int_now LL l,LL r,LL rt
LL c[maxn] , sum[maxn] ;
int main()
{
    LL n , m , l , r , x , i , temp , num ;
    while(scanf("%I64d", &n) && n)
    {
        num = 0 ;
        memset(c,0,sizeof(c));
        scanf("%I64dd", &m);
        while(m--)
        {
            scanf("%I64d %I64d %I64d", &l, &r, &x);
            c[l] += x ;
            c[r+1] -= x ;
        }
        sum[0] = 0 ;
        for(i = 1 ; i <= n ; i++)
        {
            sum[i] = sum[i-1] + c[i] ;
        }
        for(i = n-1 ; i >= 1 ; i--)
        {
            sum[i] = sum[i] + sum[i+1] ;
        }
        scanf("%I64d", &m);
        while(m--)
        {
            scanf("%I64d %I64d", &x, &l);
            if( sum[l] < x )
                num++ ;
        }
        printf("%I64d\n", num);
    }
    return 0;
}