zoj3623 Battle Ships

zoj3623 Battle Ships

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

這題比較難想到，看了題解後才恍然大悟。。這裡把時間看做容量，所打怪物的血為價值，用dp[j]表示j這個時候最多能打怪物的血量，那麼利用完全背包可以得到dp[j]=max(dp[j],dp[j-t[i]]+(j-t[i])*l[i])，其中dp[j-t[i]]是因為剛開始制造要花t[i]時間，後面指的是制造出來後的每一秒可以打多少血。

#include #include int max(int a,int b){ return a>b?a:b; } int dp[500]; int main() { int n,m=490,last,i,j; int t[50],l[50]; while(scanf(%d%d,&n,&last)!=EOF) { for(i=1;i<=n;i++){ scanf(%d%d,&t[i],&l[i]); } memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++){ for(j=t[i];j<=m;j++){ dp[j]=max(dp[j],dp[j-t[i]]+(j-t[i])*l[i]); } } for(i=1;i<=m;i++){ if(dp[i]>=last){ printf(%d ,i);break; } } } return 0; }