Battle Ships(复习泛化物品**)
Posted 小时のblog
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Battle Ships(复习泛化物品**)相关的知识,希望对你有一定的参考价值。
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 ti seconds to produce the i-th battle ship and this battle ship can make the tower loss li 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 li(1 ≤
li ≤ 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
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int f[340],protim[40],fi[40]; int n,l,i,j; int main() { while(scanf("%d%d",&n,&l)!=EOF) { memset(f,0,sizeof(f)); for(i=1;i<=n;i++) scanf("%d%d",&protim[i],&fi[i]); for(i=1;i<=l;i++) for(j=1;j<=n;j++) f[i+protim[j]]=max(f[i+protim[j]],f[i]+i*fi[j]); for( i=1;i<=340;i++) if(f[i]>=l)break; printf("%d\n",i); } return 0; }
以上是关于Battle Ships(复习泛化物品**)的主要内容,如果未能解决你的问题,请参考以下文章
CodeForces - 567D One-Dimensional Battle Ships
51 nod 1521 一维战舰 时间复杂度O(n),同 Codeforces 567D. One-Dimensional Battle Ships 有详细注释