zoj3623（更新顺序影响下的完全背包）

地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3623

Battle Ships

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 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 11 12 101 12 53 1001 103 2010 100

Sample Output

245

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;}