传送门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 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
【题目大意】有l滴血，你有n个战舰，生产每个战舰需要时间，生产后每秒攻击i点血，求耗完l滴血的最少时间。
【code】
#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;
}
时间： 2024-12-20 14:48:38