ZOJ3623 Battle Ships

ZOJ Problem Set - 3623

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

1 1
1 1
2 10
1 1
2 5
3 100
1 10
3 20
10 100

Sample Output

2
4
5

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#define N 500
using namespace std;
int dp[N];
int sum,t;
int main()
{
    int n,l;
    while(~scanf("%d%d",&n,&sum))
    {
        memset(dp,0,sizeof dp);
        int ans=10000000;
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&t,&l);
            for(int j=1;j<=400;j++)//生成后开始打的第一秒开始算
            {
                dp[j+t]=max(dp[j+t],dp[j]+l*j);//前j时间段已经打掉的长度加上第j时刻到j+t时刻这段时间能打掉的长度
                if(dp[j+t]>=sum)
                {
                    if(j+t<ans)
                    {
                        ans=j+t;
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}