HDU 4970 Killing Monsters（树状数组 || 思维）

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.[/hint]

题意：

有n座塔，每座塔的攻击范围为[l,r]，攻击力为d，有k个怪兽从这些塔前面经过，第i只怪兽初始的生命力为hp，出现的位置为x，终点为第n个格子。问最后有多少只怪兽还活着。

用数组num[x] += z; num[y+1] = -z; 然后从前向后累加一下扫一遍就 能计算出每一个格子的伤害
可以自己画图 去理解一下

code :

#include <iostream>//#include <bits/stdc++.h>#include <cstdio>#include <string.h>using namespace std;#define maxn 200000long long int sum[maxn<<1];int num[maxn<<1];int main(){    int n;    while (~scanf("%d",&n),n)    {        memset(num, 0, sizeof(num));        int m;        scanf("%d",&m);        for (int i=0; i<m; i++)        {            int x, y, z;            scanf("%d%d%d",&x,&y,&z);            num[x] += z;            num[y+1] -= z;        }        for (int i=1; i<=n ;i++) //计算每个点的伤害        {            num[i] += num[i-1];        }        for (int i=1; i<=n ;i++)  //计算第i个点之前的伤害        {            sum[i] = sum[i-1]+num[i];        }        int op;        scanf("%d",&op);        int ans=0;        long long int blood;        int now;        for (int i=0; i<op; i++)        {            scanf("%lld%d",&blood,&now);            //printf ("%lld\n",sum[n]-sum[now-1]);            if (blood>sum[n]-sum[now-1])   //比较血量和从该点到最后所需血量的大小                ans++;        }        printf ("%d\n",ans);    }    return 0;}