ZOJ 3623 Battle Ships

最新推荐文章于 2019-03-16 15:33:29 发布

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本文探讨了一个类似星际争霸的游戏策略问题，玩家需要通过合理安排不同类型的战舰生产来最小化击败敌人防御塔所需的时间。文章提供了详细的算法思路和C++代码实现。

Battle Ships

Time Limit: 2 Seconds Memory Limit: 65536 KB

Battle Ships is a new game which is similar to Star Craft. In this game, the enemy builds a defense tower, which has L longevity. The player has a military factory, which can produce N kinds of battle ships. The factory takes ti seconds to produce the i-th battle ship and this battle ship can make the tower loss li longevity every second when it has been produced. If the longevity of the tower lower than or equal to 0, the player wins. Notice that at each time, the factory can choose only one kind of battle ships to produce or do nothing. And producing more than one battle ships of the same kind is acceptable.

Your job is to find out the minimum time the player should spend to win the game.

Input

There are multiple test cases.
The first line of each case contains two integers N(1 ≤ N ≤ 30) and L(1 ≤ L ≤ 330), N is the number of the kinds of Battle Ships, L is the longevity of the Defense Tower. Then the following N lines, each line contains two integers t i(1 ≤ t i ≤ 20) and li(1 ≤ li ≤ 330) indicating the produce time and the lethality of the i-th kind Battle Ships.

Output

Output one line for each test case. An integer indicating the minimum time the player should spend to win the game.

Sample Input

Sample Output

2
4
5

题意：敌方防御塔有L的生命值，我方能建造n种飞船，建造时间为t，建造完成之后每秒能够对敌方造成l点伤害，问击败敌方最少需要多少时间。

思路：将时间作为背包容量来进行背包，那么选择一种飞船的话，就相当于在最开始的时候花了t的时间建造，之后每秒都会造成l点伤害，所以dp[j]=max(dp[j],dp[j-t[i]]+(j-t[i])*l)，dp[j]表示时间为j的时候最大伤害量。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
const int inf=0x3f3f3f3f;
int dp[350],c[35],w[30];
int main()
{
    int n,l,i,j;
    while(scanf("%d%d",&n,&l)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d%d",&c[i],&w[i]);
        memset(dp,0,sizeof(dp));
        int temp=inf;
        for(i=1;i<=n;i++)
        {
            for(j=c[i];j;j++)
            {
                dp[j]=max(dp[j],dp[j-c[i]]+(j-c[i])*w[i]);
                if(dp[j]>=l) break;
            }
            temp=min(temp,j);
        }
        printf("%d\n",temp);
    }
    return 0;
}