HDU - 4970 Killing Monsters（思维）

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Killing Monsters

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2517    Accepted Submission(s): 944

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.

清明节打了一天的组队赛，本弱猪身心俱疲，又一次真真切切的感觉到自己弱的掉渣渣。一名舍友长水痘痘回家了（=。=！都多大了，老兄，关键是谁早上叫我起床啊）

出来题后，分分钟看到J题被A掉。excuse me？？读了J题，一个ACM队长追心仪女生（=。=！然而本猪毫无头绪）猪友村庄第一次读题这么快，一眼看到这个题，听到大佬在说“线段树”。（=。=！这，就触碰到我知识盲区了）然并卵。

本猪这里没用结构体（其实都一样啦），只要计算出每一个Block能够攻击多少滴血，然后后序循环一变，然后对于每一个monster的血量是否小于这个block的攻击血量判断就OK了。由于之前的教训，本猪毅然用scanf，提交后焦急等待，WA了，发现没写清空member，满怀欣喜的再次提交，TLE（=。=！）读题，10的18次方，鼓起勇气再次提交。AC！！(罚时，本猪心里卧槽卧槽的！！！)

附上AC代码：

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll block_attack[maxn];
int N,M;
ll l,r,d;
int k;
ll hi,xi;

int main()
{
    while(scanf("%d",&N) != EOF && N)
    {
        memset(block_attack, 0, sizeof(block_attack));
        scanf("%d",&M);
        while(M--)
        {
            scanf("%lld%lld%lld",&l,&r,&d);
            block_attack[l] += d;
            block_attack[r+1] -= d;
        }
        for(int i = 1; i < N; i++)
            block_attack[i] += block_attack[i-1];
        for(int i = N - 2; i >= 0; i--)
            block_attack[i] += block_attack[i+1];
        int ans = 0;
        scanf("%d",&k);
        while(k--)
        {
            scanf("%lld%lld",&hi,&xi);
            if(block_attack[xi] < hi)ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}