ZOJ Problem Set - 3623 Battle Ships

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

---

Author: FU, Yujun

            
            
            
                Contest: ZOJ Monthly, July 2012

            
            
 //dp[i]代表时间为i时最多可以打掉的血
//然后找到第一个dp[i]>=L,那么就是最短的时间
//采用的是用当前状态去更新后面状态
//对于，j+T[i]时间段内，
//dp[j]更新dp[j+T[i]]，表示一开始的T[i]秒是在建武器i
//在j的时间段内武器i一直都在工作，然后J时间段内套用j+T[i]时间段内的含义.
//所以该动规方法是可行的
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int N,L;
int T[34],l[34];
int dp[334];
int main()
{
    int i,j;
    while(scanf("%d%d",&N,&L)!=EOF)
    {
        for(i=1;i<=N;i++)
         scanf("%d%d",&T[i],&l[i]);
        memset(dp,0,sizeof(dp));
     for(j=1;j<=L;j++)
        for(i=1;i<=N;i++)
          dp[j+T[i]]=max(dp[j]+l[i]*j,dp[j+T[i]]);
        for(i=1;i<=330;i++)
         if(dp[i]>=L)
          break;
        printf("%d\n",i);
    }
    return 0;
}