1056 Mice and Rice (25)

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.
First the playing order is randomly decided for NP​ programmers. Then every NG​ programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every NG​ winners are then grouped in the next match until a final winner is determined.
For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.
Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: NP​ and NG​ (≤1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG​ mice at the end of the player's list, then all the mice left will be put into the last group. The second line contains NP​ distinct non-negative numbers Wi​ (i=0,⋯,NP​−1) where each Wi​ is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,⋯,NP​−1 (assume that the programmers are numbered from 0 to NP​−1). All the numbers in a line are separated by a space.
Output Specification:

For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.
Sample Input:

2. 11 3
3. 25 18 0 46 37 3 19 22 57 56 10
4. 6 0 8 7 10 5 9 1 4 2 3

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Sample Output:

2. 5 5 5 2 5 5 5 3 1 3 5

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标题大意：np为老鼠的数目，ng为每组最多g个老鼠。先给出np个老鼠的重量，再给出老鼠的初始序次（第i名的老鼠是第j号，j从0开始）。每ng个老鼠分为一组（末了不敷np的单独为1组），对于每组老鼠，选出最重的谁人，晋级下一轮角逐，未晋级的排名为队伍数目+1，再依次再以np个老鼠一组分类，然后选出重量最大的，直到只剩下一只老鼠，排名为1。这个排名是按照原输入老鼠的序次输出老鼠的排名。
分析：模拟角逐即可。注意第一次角逐是按照给定的序次举行，之后的角逐是按照每次晋级老鼠的位置举行。

2. #include<algorithm>
3. #include <iostream>
4. #include  <cstdlib>
5. #include  <cstring>
6. #include   <string>
7. #include   <vector>
8. #include   <cstdio>
9. #include    <queue>
10. #include    <stack>
11. #include    <ctime>
12. #include    <cmath>
13. #include      <map>
14. #include      <set>
15. #define INF 0xffffffff
16. #define db1(x) cout<<#x<<"="<<(x)<<endl
17. #define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
18. #define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
19. #define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
20. using namespace std;
22. typedef struct
23. {
24.     int weight,ind;
25. }mice;
27. int main(void)
28. {
29.     #ifdef test
30.     freopen("in.txt","r",stdin);
31.     //freopen("in.txt","w",stdout);
32.     clock_t start=clock();
33.     #endif //test
35.     int np,ng;scanf("%d%d",&np,&ng);
36.     mice num[np],cnt[np];
37.     int ans[np]={0},que[np]={0};
38.     int t=np;
39.     for(int i=0;i<np;++i)
40.         scanf("%d",&num[i].weight),num[i].ind=i;
41.     for(int i=0;i<np;++i)
42.         scanf("%d",&que[i]);
44.     int tt=0;
45.     int ind=0,val;
46.     if(t%ng!=0)val=t/ng+2;
47.     else val=t/ng+1;
48.     for(int i=0;i<t;i=i+ng)
49.     {
50.         int temp=num[que[i]].weight,index=i,r=min(i+ng,t);
51.         for(int j=i+1;j<r;++j)
52.         {
53.             if(num[que[j]].weight>temp)temp=num[que[j]].weight,index=j;
54.         }
55.         for(int j=i;j<r;++j)
56.         {
57.             if(j!=index)ans[num[que[j]].ind]=val;
58.         }
59.         cnt[tt]=num[que[index]];
60.         tt++;
61.     }
62.     t=tt,tt=0;
63.     for(int i=0;i<t;++i)
64.         num[i]=cnt[i];
66.     while(t>1)
67.     {
68.         ind=0;
69.         if(t%ng!=0)val=t/ng+2;
70.         else val=t/ng+1;
72.         for(int i=0;i<t;i=i+ng)
73.         {
74.             int temp=num[i].weight,index=i,r=min(i+ng,t);
75.             for(int j=i+1;j<r;++j)
76.             {
77.                 if(num[j].weight>temp)temp=num[j].weight,index=j;
78.             }
79.             for(int j=i;j<r;++j)
80.             {
81.                 if(j!=index)ans[num[j].ind]=val;
82.             }
83.             cnt[tt]=num[index];
84.             tt++;
85.         }
86.         for(int i=0;i<tt;++i)
87.             num[i]=cnt[i];
88.         t=tt,tt=0;
89.     }
90.     ans[num[0].ind]=1;
91.     for(int i=0;i<np;++i)
92.     {
93.         if(!i)printf("%d",ans[i]);
94.         else printf(" %d",ans[i]);
95.     }
96.     printf("\n");
98.     #ifdef test
99.     clockid_t end=clock();
100.     double endtime=(double)(end-start)/CLOCKS_PER_SEC;
101.     printf("\n\n\n\n\n");
102.     cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
103.     cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
104.     #endif //test
105.     return 0;
106. }

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