HDU4970 Killing Monsters （2014多校第九场第11题）

Killing Monsters

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 491    Accepted Submission(s): 280

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.
 

 

Author

SYSU

 

Source

2014 Multi-University Training Contest 9

 

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <queue>#include <numeric>#include <list>#include <functional>using namespace std; #define eps 1e-8#define LL long long#define ULL unsigned long long#define mod 1000000007#define MP make_pair#define PII pair<int, int>#define INF 0x3f3f3f3fLL attack[100005];int main(){int n, m, k;intl, r, x;LL hp;//freopen("1011.in", "r", stdin);//freopen("out.txt", "w", stdout);while (~scanf("%d%d", &n, &m), n){memset(attack, 0, sizeof(attack));//scanf("%d", &m);for (int i = 0; i < m; ++i){scanf("%d%d%d", &l, &r, &x);attack[l] += x; attack[r + 1] -= x;}for (int i = 1; i <= n; ++i){attack[i] += attack[i-1];}for (int i = n - 1; i >= 1; --i){attack[i] += attack[i+1];}scanf("%d", &k);int ans = 0;for (int i = 0; i < k; ++i){scanf("%I64d%d", &hp, &l);if (hp > attack[l]) ++ans;}//printf("%I64d\n", ans); 被HDU输出方式搞萌，I64d wa, lld AC；cout << ans << endl;}return 0;}

题解：刚开始是写的线段树，然后T了，所以要你机智的过。机智的姿势是：读入l,r,x,表示range[l, r]每次攻击x的伤害，另a[l] += x, a[r + 1] -= x, 这样子，读完后，从左往右a[i] += a[i-1]就会更新掉每个地方每次总共的攻击值，再从右往左a[i] += a[i-1]， 就能得出第i个block到第N个block受到的总伤害。