Jobdu 1494:Dota - KingEasternSun的专栏 - 博客频道 - CSDN.NET


九度笔记之 1494:Dota - KingEasternSun的专栏 - 博客频道 - CSDN.NET
大家都知道在dota游戏中,装备是对于英雄来说十分重要的要素。
英雄们不仅可以购买单个的装备,甚至某些特定的装备组合能够合成更强的装备。
为了简化问题,我们将每个装备对于英雄的功能抽象为一个整数:价值。同时,如上所说,一些特定的装备可以用来合成更强的装备,玩家会因此获得除原装备价值外额外的价值。
给定玩家现有的金钱数,每个装备的价格和其对应的价值,以及装备合成的信息。输出,其能获得的最大价值数。
注意:每件装备只能参与合成一件合成装备(即原装备参与合成后得到合成后的新装备,原装备消失),除非一次购买多个该种装备。
输入:
输入包含多组测试数据,每组测试数据第一行为三个整数n,m,g(1<=n<=100),(0<=m<=100),(1<=g<=1000)。
分别表示存在的装备总数n,存在的装备合成关系数m,和英雄所有的金钱g。
接下去n行,每行两个整数(p,v)(1<=p<=100,1<=v<=100)分别表示该件装备的价格p,和其自身的价值v。
每组测试数据的最后为m行数据,每行由两部分组成。开头一个整数t(1<=t<=n),表示该组合成关系中需要t件装备,下去紧跟t件装备编号(按照装备在输入中的顺序从1到n编号),表示参与合成该装备的装备编号,最后一个数s(1<=s<=1000),代表这t件物品合成道具后获得的额外价值。
输出:
对于每组测试数据,输出一个整数,代表英雄可以获得的最大价值。
样例输入:
3 1 100
100 20
50 9
50 9
2 2 3 1
3 1 100
100 20
50 9
50 9
2 2 3 3
样例输出:
20
21
http://blog.csdn.net/u013027996/article/details/26989653
装备是没有购买上限的, 这就简单多了
2. 假如给定的装备只能取一件, 装备之间又能合成新装备, 那么这道题就难了
完全背包,内层循环是从小到大。注意合成值也要单独得到花费和价值。
  1.     public static void main(String[] args) throws Exception {  
  2.         StreamTokenizer st = new StreamTokenizer(new BufferedReader(  
  3.                 new InputStreamReader(System.in)));  
  4.         while (st.nextToken() != StreamTokenizer.TT_EOF) {  
  5.             int n = (int) st.nval;  
  6.             st.nextToken();  
  7.             int m = (int) st.nval;  
  8.             st.nextToken();  
  9.             int g = (int) st.nval;  
  10.             int size = n + 1 + m;  
  11.             int p[] = new int[size];  
  12.             int v[] = new int[size];  
  13.             for (int i = 1; i < n + 1; i++) {  
  14.                 st.nextToken();  
  15.                 p[i] = (int) st.nval;  
  16.                 st.nextToken();  
  17.                 v[i] = (int) st.nval;  
  18.             }  
  19.             for (int i = n + 1; i < size; i++) {  
  20.                 st.nextToken();  
  21.                 int t = (int) st.nval;  
  22.                 for (int j = 1; j < t + 1; j++) {  
  23.                     st.nextToken();  
  24.                     int seq = (int) st.nval;  
  25.                     p[i] += p[seq];  
  26.                     v[i] += v[seq];  
  27.                 }  
  28.                 st.nextToken();  
  29.                 int s = (int) st.nval;  
  30.                 v[i] += s;  
  31.             }  
  32.             int dp[] = new int[g + 1];  
  33.             for (int i = 1; i < size; i++) {  
  34.                 for (int j = p[i]; j <= g; j++) {  
  35.                     dp[j] = Math.max(dp[j], dp[j - p[i]] + v[i]);  
  36.                 }  
  37.             }  
  38.             System.out.println(dp[g]);  
  39.         }  
  40.     }  
  41. }  

http://blog.csdn.net/wdy_yx/article/details/9815631
我们把合成的装备看作单独的一件装备,计算合成装备的价格(合成所需装备价格之和),合成装备的价值(合成所需装备价值之和 加上 额外价值)
  1. for(int i = num_weapons;i<num_weapons+num_merge;i++){  
  2.     int t=0;  
  3.     std::cin>>t;  
  4.     price[i]=0;  
  5.     value[i]=0;  
  6.     for(int j = 0;j<t;j++){  
  7.         int id_weapon;  
  8.         std::cin>>id_weapon;  
  9.         price[i]+=price[id_weapon-1];  
  10.         value[i]+=value[id_weapon-1];  
  11.     }  
  12.     int s = 0;  
  13.     std::cin>>s;  
  14.     value[i]+=s;  
  15. }  
2.这道题和其它背包问题不同的在于 装备的数量并不是固定的,玩家可以重复购买多个同一种装备,包括合成装备,只要钱够就行。所以在动态规划更新dp的时候,就是从低到高更新。
  1. void dota(int num_weapons,int num_merge,int gold){  
  2.     int *price = new int[num_weapons+num_merge];  
  3.     int *value = new int[num_weapons+num_merge];  
  4.     int *maxvalue = new int[gold+1];  
  5.     for(int i = 0;i<num_weapons;i++){  
  6.         std::cin>>price[i]>>value[i];  
  7.     }  
  8.     for(int i = num_weapons;i<num_weapons+num_merge;i++){  
  9.         int t=0;  
  10.         std::cin>>t;  
  11.         price[i]=0;  
  12.         value[i]=0;  
  13.         for(int j = 0;j<t;j++){  
  14.             int id_weapon;  
  15.             std::cin>>id_weapon;  
  16.             price[i]+=price[id_weapon-1];  
  17.             value[i]+=value[id_weapon-1];  
  18.         }  
  19.         int s = 0;  
  20.         std::cin>>s;  
  21.         value[i]+=s;  
  22.     }  
  23.    
  24.     for(int i = 0;i<gold+1;i++){  
  25.         maxvalue[i]=0;  
  26.     }  
  27.     for(int i = 0;i<num_weapons+num_merge;i++){  
  28.         for(int j = price[i];j<gold+1;j++){  
  29.             maxvalue[j] = std::max(maxvalue[j],maxvalue[j-price[i]]+value[i]);  
  30.         }  
  31.     }  
  32.     std::cout<< maxvalue[gold]<<std::endl;  
  33.     delete []price;  
  34.     delete []value;  
  35.     delete []maxvalue;  
  36.    
  37. }  
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