ZOJ3623: Battle Ships（类完全背包）

最新推荐文章于 2019-03-16 15:33:29 发布

junior19

最新推荐文章于 2019-03-16 15:33:29 发布

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分类专栏：背包问题

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本文介绍了一款类似星际争霸的游戏——战舰游戏，并详细解析了如何通过动态规划算法求解玩家赢得游戏所需的最短时间。游戏设定玩家拥有一座能够制造不同种类战舰的军工厂，每种战舰有不同的生产时间和攻击力。

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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 t_i seconds to produce the i-th battle ship and this battle ship can make the tower loss l_i 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 l_i(1 ≤ l_i ≤ 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

Author: FU, Yujun
Contest: ZOJ Monthly, July 2012

dp[i]表示i时间内能制造的最大伤害，dp[i] = max( dp[i], dp[i-time[j]] + (i-time[j])*value[j] )，可以理解为dp[i-time[j]]的事件保持不变，而在事件前面增加time[j]的时间制造第j种船，该船在接下来i-time[j]时间内造成额外的伤害(i-time[j])*value[j]。

# include <stdio.h>
# include <string.h>
# define INF 0x3f3f3f3f
int dp[333], v[31],c[31];
int main()
{
    int n, t, i, j;
    while(~scanf("%d%d",&n,&t))
    {
        memset(dp, 0, sizeof(dp));
        int cnt = INF;
        for(i=0; i<n; ++i)
            scanf("%d%d",&c[i],&v[i]);
        for(i=0; i<n; ++i)
        {
            for(j=c[i]; j; ++j)
            {
                if(dp[j-c[i]]+v[i]*(j-c[i]) > dp[j])
                    dp[j] = dp[j-c[i]]+v[i]*(j-c[i]);
                if(dp[j] >= t)
                    break;
            }
            if(j < cnt)
                cnt = j;
        }
        printf("%d\n",cnt);
    }
    return 0;
}