2017百度之星资格赛 1003：度度熊与邪恶大魔王（DP）

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度度熊与邪恶大魔王

Accepts: 3021

Submissions: 18787

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Problem Description

度度熊为了拯救可爱的公主，于是与邪恶大魔王战斗起来。 邪恶大魔王的麾下有n个怪兽，每个怪兽有a[i]的生命值，以及b[i]的防御力。 度度熊一共拥有m种攻击方式，第i种攻击方式，需要消耗k[i]的晶石，造成p[i]点伤害。 当然，如果度度熊使用第i个技能打在第j个怪兽上面的话，会使得第j个怪兽的生命值减少p[i]-b[j]，当然如果伤害小于防御，那么攻击就不会奏效。 如果怪兽的生命值降为0或以下，那么怪兽就会被消灭。 当然每个技能都可以使用无限次。 请问度度熊最少携带多少晶石，就可以消灭所有的怪兽。

Input

本题包含若干组测试数据。 第一行两个整数n，m,表示有n个怪兽，m种技能。 接下来n行，每行两个整数，a[i],b[i]，分别表示怪兽的生命值和防御力。 再接下来m行，每行两个整数k[i]和p[i]，分别表示技能的消耗晶石数目和技能的伤害值。 数据范围: 1<=n<=100000 1<=m<=1000 1<=a[i]<=1000 0<=b[i]<=10 0<=k[i]<=100000 0<=p[i]<=1000

Output

对于每组测试数据，输出最小的晶石消耗数量，如果不能击败所有的怪兽，输出-1

Sample Input

Copy

1 2
3 5
7 10
6 8
1 2
3 5
10 7
8 6

Sample Output

Copy

6
18

http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=774&pid=1003

就是一个完全背包，每个怪兽可以分开算，最后把答案加起来就是总的答案；由于怪兽的防御值范围很小，可以预处理出每一组技能下打死不同生命、不同防御值的怪物消耗的最少晶石，对于每个怪物直接在其中取答案，比对于每个怪物求一次要快很多。

自己的代码&讲解：（太丑 了，什么一直求到2*maxa也没谁了。。）

 /*
 ans[i][j][b]表示防御力为b的怪物
 表示用前i种技能，造成恰好j的伤害消耗的最小晶石
 ans[i][j][b]=min{ans[i-1][j-x*(p[i]-b)]+x*k[i]}
 x表示第i种技能取x个(0<=x*(p[i]-b)<=j)
 优化：http://blog.csdn.net/wumuzi520/article/details/7014830
 ans[i][j][b]=min(ans[i-1][j][b],ans[i][j-(p[i]-b)][b]+k[i]【j>=p[i]-b】)
 =>ans[j][b]=min(ans[j][b],ans[j-(p[i]-b)][b]+k[i] **i从小到大
 */
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long LL;
 LL n,m;
 LL maxb,maxp,bx,maxa;
 LL a[],b[];
 LL ans[][];
 LL k[],p[],ans1;
 int main()
 {
     LL i,j,t,ze=(long long);
     while(scanf("%lld%lld",&n,&m)==)
     {
         maxb=;maxp=;maxa=;
         for(i=;i<=n;i++)
         {
             scanf("%lld%lld",&a[i],&b[i]);
             maxb=max(maxb,b[i]);
             maxa=max(maxa,a[i]);
         }
         for(i=;i<=m;i++)
         {
             scanf("%lld%lld",&k[i],&p[i]);
             maxp=max(maxp,p[i]);
         }
         maxa=max(maxa,maxp);
         /*
         去掉以上这句，会让类似
         1 1
         45 10
         4164 327
         的数据得出错误结果
         */
         if(maxb>=maxp)
         {
             printf("-1\n");
             continue;
         }
         memset(ans,0x3f,sizeof(ans));
         memset(ans[],,sizeof(ans[]));
         for(bx=;bx<=maxb;bx++)
         {
             for(i=;i<=m;i++)
                 for(j=max(ze,p[i]-bx);j<=*maxa;j++)
             //    {
                     //if(j>=p[i]-bx)
                         ans[j][bx]=min(ans[j][bx],ans[j-p[i]+bx][bx]+k[i]);
             //    }
             t=0x3f3f3f3f;
             for(j=*maxa;j>=;j--)
             {
                 t=min(t,ans[j][bx]);
                 ans[j][bx]=min(ans[j][bx],t);
             }
         }
         ans1=;
         for(i=;i<=n;i++)
             ans1+=ans[a[i]][b[i]];
         printf("%lld\n",ans1);

     }
     return ;
 }

题解：2017百度之星资格赛 1003：度度熊与邪恶大魔王（DP）