TOJ 4120 Zombies VS Plants

最新推荐文章于 2023-02-25 17:02:00 发布

转载最新推荐文章于 2023-02-25 17:02:00 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/Basasuya/p/8433779.html

本文解析了一个基于游戏《植物大战僵尸》改编的算法问题——ZombiesVSPlants。玩家扮演僵尸管理者，目标是在每行放置足够的僵尸以吃掉所有植物，并计算所需的最小总生命值。文章提供了完整的 C++ 代码实现，重点处理了输入中可能出现的大量空白字符的问题。

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链接：http://acm.tju.edu.cn/toj/showp4120.html4120. Zombies VS Plants

Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 268 Accepted Runs: 36

tmeteorj recently is interested in a game Plants VS Zombies like following picture.

And his brain hole is opened so he want to you to solve this problem.
Your role is zombies' manager, and your task is place enough zombies to eat all plants on each line.
The rules are as follows:
1.Each zombies has health point from 1 to infinity.
2.Each plants has its attack point which we will tell you and will not greater than 1000.
3.All plants are so afraid of zombies, so if a plant's attack point is not greater than it's line's zombie's HP, this plant will be scared silly. So its attack point will become zero.
4.Zombies eat plants from right to the left.
5.You can place any number of zombies on each line.
Your task is to calculate the minimal total zombies' HP you need to place in order to eat all plants.

INPUT

First line, there will a number T(T≤10), means the test case of this problem.
After this, there will be a number n(n≤100), means there are n lines in the yard.
Each line contains several plants' attack points. The plants' number will not greater than 100.
There may be some blanks in anywhere.

OUTPUT

For each case, print the minimal HP for the question.

Sample Input


3
2
1 2 3 4 5
2 3 4 5
1
1 2 3 4 5
2
1 2 3 4 5 6
1 2

Sample Output


10
5
8

/*
遇见过的最坑的题目，主要坑点在与输入会在各处出现空格
听说还可能出现一行1500空格....
所以得用好getchar，这是关键
*/ 
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<map>
using namespace std;
int main()
{
	int i,j,k,zhi,fu,flag,N,T;
	char ch;
	int sum,mx ;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&N);
		while(getchar()!='\n');
		sum=0;
		for(int i=1;i<=N;i++){
			flag=0;
			zhi=0;
			fu=1;
			mx=1;
			while(1){
				ch=getchar();
				if(ch=='-'){
					flag=1;
					fu=-1;
				}
				else if(ch==' '){
					mx=max(mx,fu*zhi);
					fu=1;
					zhi=0;
				}
				else if(ch=='\n'){
					if(flag){
						mx=max(mx,fu*zhi);
						fu=1;
						zhi=0;
					}
					else mx=0;
					break;
				}
				else{
					zhi=zhi*10+ch-'0';
					flag=1;
				}
			}
			sum+=mx;
		}
		printf("%d\n",sum);
	}
	return 0;
}

转载于:https://www.cnblogs.com/Basasuya/p/8433779.html