HDU 2159 Fate


Problem - 2159
最近xhd正在玩一款叫做FATE的游戏,为了得到极品装备,xhd在不停的杀怪做任务。久而久之xhd开始对杀怪产生的厌恶感,但又不得不通过杀怪来升完这最后一级。现在的问题是,xhd升掉最后一级还需n的经验值,xhd还留有m的忍耐度,每杀一个怪xhd会得到相应的经验,并减掉相应的忍耐度。当忍耐度降到0或者0以下时,xhd就不会玩这游戏。xhd还说了他最多只杀s只怪。请问他能升掉这最后一级吗?

Input
输入数据有多组,对于每组数据第一行输入n,m,k,s(0 < n,m,k,s < 100)四个正整数。分别表示还需的经验值,保留的忍耐度,怪的种数和最多的杀怪数。接下来输入k行数据。每行数据输入两个正整数a,b(0 < a,b < 20);分别表示杀掉一只这种怪xhd会得到的经验值和会减掉的忍耐度。(每种怪都有无数个)

Output
输出升完这级还能保留的最大忍耐度,如果无法升完这级输出-1。

Sample Input
10 10 1 10 1 1 10 10 1 9 1 1 9 10 2 10 1 1 2 2
Sample Output
0 -1 1
限制条件: 1.忍耐度 m   2.杀怪个数 s
http://www.cnblogs.com/dongsheng/archive/2012/08/22/2651614.html
 4 int dp[101][101];      //dp[i][j] 表示消耗i的忍耐度和杀j个怪物得到的最大经验值
 5 struct node
 6 {
 7     int e;     //经验值
 8     int r;     //忍耐度
 9 }a[101];
10 
11 int main()
12 {
13     int n,m,k,s,i,j,t;
14     while(scanf("%d%d%d%d",&n,&m,&k,&s)!=EOF)
15     {
16         for(i=1;i<=k;++i)
17             scanf("%d%d",&a[i].e,&a[i].r);
18         memset(dp,0,sizeof(dp));
19         for(i=1;i<=k;++i)   //k表示怪物种类---对不同怪物遍历一遍
20             for(j=a[i].r;j<=m;++j)  //m表示保留的忍耐度
21                 for(t=1;t<=s;++t)    // s表示杀的怪物数
22                 {
23                     if(dp[j][t]<dp[j-a[i].r][t-1]+a[i].e)
24                     {
25                         dp[j][t]=dp[j-a[i].r][t-1]+a[i].e;
26                     }
27                 }
28         if(dp[m][s]>=n)     //表示能过升级
29         {
30                 for(i=0;i<=m;++i)   //寻找能够升级所消耗的最小忍耐度,只用找消耗相同忍耐度的情况下,令杀怪数量最多,
31                     if(dp[i][s]>=n) //那么d[i][s]一定是消耗i忍耐度的情况下,获得的最大经验值
32                     {
33                         printf("%d\n",m-i);
34                         break;
35                     }
36         }
37         else
38             printf("-1\n");
39     }
40     return 0;
41 }
https://www.jianshu.com/p/1f2aa01a7149
    while(scanf("%d%d%d%d",&n,&m,&k,&s)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=k;i++)
        {
            scanf("%d%d",exp+i,cost+i);
        }
        for(int i=1;i<=k;i++)
        {
            for(int j=1;j<=s;j++)
            {
                for(int q=cost[i];q<=m;q++)
                {
                    dp[j][q]=Max(dp[j][q],dp[j-1][q-cost[i]]+exp[i]);
                }
            }
        }
        if(dp[s][m]<n)
        {
            printf("-1\n");
        }
        else
        {
            int tmp=0;
            for(int i=1;i<=m;i++)
            {
                if(dp[s][i]>=n)
                {
                    tmp=i;
                    break;
                }
            }
            printf("%d\n",m-tmp);
        }
    }

}
http://blog.csdn.net/u013476556/article/details/38349831
dp[j][k]的意思的是  在忍耐度为j,杀了k个怪的状态下所对应的经验值。
dp[j-wei[i]][k-1] + val[i];在忍耐度为  j - wei[i],杀了k-1个怪的状态下所对应的经验值。
第三个for循环正序逆序都会AC  。。
  1.    while(scanf("%d%d%d%d",&n,&m,&k,&s)!=EOF)///经验值 忍耐度 种类 杀怪数  
  2.     {  
  3.         for(int i=0; i<k; i++)  
  4.         {  
  5.             scanf("%d%d",&val[i],&wei[i]);  
  6.         }  
  7.         memset(dp,0,sizeof(dp));  
  8.         for(int i = 0; i < k; i++) ///种类  
  9.         {  
  10.             for(int j = wei[i]; j <= m; j++) ///忍耐度  
  11.             {  
  12.                 for(int k = s; k >= 1; k--) ///杀怪数  
  13.                 {  
  14.                     dp[j][k] = Max(dp[j][k],dp[j-wei[i]][k-1] + val[i]);///在忍耐度为j,杀了k个怪的状态下所对应的经验值  
  15.                 }  
  16.             }  
  17.         }  
  18.         int i;  
  19.         for (i =0 ;i <= m; i++)  
  20.         {  
  21.             if (dp[i][s] >= n)  
  22.                 break;  
  23.         }  
  24.         if (i > m)  
  25.             printf ("-1\n");  
  26.         else printf ("%d\n",m - i);  
  27.     }  

http://www.programerhome.com/?p=4537
思路:这题是一道典型的二维完全背包题,背包内所要储存的是经验,所以背包的容量便以忍耐度与杀怪数作为标准,每次得到背包价值的最大数与升级所需的经验作比较,能够升级就退出。
    while(cin>>n>>m>>k>>s)//经验值,忍耐度,怪的种数和最多杀怪数
    {
        for(int i=1; i<=k; ++i)
            cin>>a[i].v>>a[i].w;

        memset(dp, 0, sizeof(dp));

        for(x=1; x<=m; x++)
        {
            for(y=1; y<=k; ++y)
                for(z=1; z<=s; ++z)
                {
                    int st=1;
                    while(st*a[y].w<=x&&st<=z)
                    {
                        dp[x][z]=max(dp[x-st*a[y].w][z-st]+st*a[y].v,dp[x][z]);
                        st++;
                    }

                }
            if(dp[x][s]>=n)
                break;
        }
        if(x>m)
            cout<<-1<<endl;
        else
            cout<<m-x<<endl;

    }
http://blog.csdn.net/zfz1015/article/details/7854788


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