ZOJ3623:Battle Ships(全然背包)

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_iseconds 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

题意：

对方有L滴血，我们有n种船能够选择。每种船建造时间为t，建好后每秒对敌方造成l点伤害，问最少多少时间能干掉对方

思路：

以时间为容量来进行背包

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define inf 1<<30
int dp[10000],t[100],l[100];

int main()
{
    int n,m,i,j,sum = 6600,ans;
    while(~scanf("%d%d",&n,&m))
    {
        for(i = 0; i<n; i++)
            scanf("%d%d",&t[i],&l[i]);
        memset(dp,0,sizeof(dp));
        for(j = 0; j<333; j++)
        {
            for(i = 0; i<n; i++)
            {
                dp[j+t[i]] = max(dp[j+t[i]],dp[j]+j*l[i]);
            }
        }
        ans = inf;
        for(i = 0; i<333; i++)
        {
            if(dp[i]>=m)
            {
                ans=min(ans,i);
            }
        }
        printf("%d\n",ans);
    }

    return 0;
}