zoj3623(更新顺序影响下的完全背包)
2013-11-07 21:25
411 查看
地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3623
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 produceN 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.
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.
题意:打一个血量为v的小怪兽,可以建n种塔,每种塔耗时ti秒每秒打怪ai点血。最少要多少秒可以杀死小怪兽。
思路:以时间为容量,伤害的总血量为价值。因为塔可以在建造好后每一秒都造成伤害,所以建造同种塔的情况下,先后顺序会影响最终结果。
代码:
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 produceN 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
题意:打一个血量为v的小怪兽,可以建n种塔,每种塔耗时ti秒每秒打怪ai点血。最少要多少秒可以杀死小怪兽。
思路:以时间为容量,伤害的总血量为价值。因为塔可以在建造好后每一秒都造成伤害,所以建造同种塔的情况下,先后顺序会影响最终结果。
代码:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int dp[700],vol[40],val[40]; int main() { int n,v,i,j,ans; while(scanf("%d%d",&n,&v)>0) { memset(dp,0,sizeof(dp)); for(i=0; i<n; i++) scanf("%d%d",&vol[i],&val[i]); ans=1100; for(i=0; i<n; i++) { for(j=vol[i]; ; j++) { if(dp[j-vol[i]]+(j-vol[i])*val[i]>dp[j]) dp[j]=dp[j-vol[i]]+(j-vol[i])*val[i]; //(j-vol[i])*val[i]只先建造种类为i的塔,然后再加上其后时间内的最优方案 if(dp[j]>=v) break; } if(j<ans) ans=j; } printf("%d\n",ans); } return 0; }
相关文章推荐
- zoj3623 Battle Ships ——完全背包?简单DP!|| 泛化背包
- hdu 3127 完全背包 二维限制条件 放置顺序相关性
- soj 3300 Stockholm Coins(完全背包的更新次数)
- ZOJ3623:Battle Ships(完全背包)
- zoj3623-完全背包(向前插入)
- 01背包与完全背包就差一个顺序
- ZOJ3623: Battle Ships(类完全背包)
- 关于01背包和完全背包二重循环的顺序(前人之技,后人惊叹)
- ZOJ3623 Battle Ships (完全背包)
- 背包DP(01背包问题,完全背包问题) 经典题
- POJ 2063 Investment(完全背包)
- 01,完全,多重背包
- HDU-1114 完全背包+恰好装满问题
- 01背包+完全背包问题
- hdu2159二维完全背包
- windows xp,windows 7下ubuntu11.10硬盘usb安装, 系统独立,互不影响重装,更新。
- CODEFORCES 189A(完全背包完全装满)
- 独家 | 英伟达回应禁令:研究人员放心用,不更新驱动就没影响
- 百度搜索中关键词顺序对搜索结果排序的影响
- 完全免费的Virtual PC和更新的SP1