PAT 1056. Mice and Rice (25)
来源:互联网 发布:国内网络加速器 编辑:程序博客网 时间:2024/05/29 18:50
http://pat.zju.edu.cn/contests/pat-a-practise/1056
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 NGwinners 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 Wiis 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:11 325 18 0 46 37 3 19 22 57 56 106 0 8 7 10 5 9 1 4 2 3Sample Output:
5 5 5 2 5 5 5 3 1 3 5
#include <cstdio>#include <queue>using namespace std;#define MAX 1001int w[MAX];int ans[MAX];int main(){//freopen("G:\\input.in", "r", stdin);int n, m;scanf("%d%d", &n, &m);for (int i = 0; i < n; i++)scanf("%d", &w[i]);queue<int> order;int x;for (int i = 0; i < n; i++){scanf("%d", &x);order.push(x);}while (order.size() != 1){queue<int> tmp;int groupNum = order.size() / m + (order.size()%m == 0 ? 0:1);for (int i = 0; i < groupNum; i++){int max = -1, maxIndex = -1;for (int j = i*m; j < i*m + m && order.size()!= 0; j++){int t = order.front();ans[t] = groupNum + 1;order.pop();if (w[t] > max){max = w[t];maxIndex = t;}}tmp.push(maxIndex);}order = tmp;}ans[order.front()] = 1;for (int i = 0; i < n; i++)printf("%d%c", ans[i], (i-n+1? ' ': '\n'));return 0;}
- 1056. Mice and Rice (25)-PAT
- PAT 1056. Mice and Rice (25)
- PAT A 1056. Mice and Rice (25)
- PAT 1056. Mice and Rice (25)
- PAT 1056. Mice and Rice (25)
- 【PAT】1056. Mice and Rice (25)
- PAT(A) - 1056. Mice and Rice (25)
- 【PAT甲级】1056. Mice and Rice (25)
- 1056. Mice and Rice (25)PAT甲级
- PAT甲级1056. Mice and Rice (25)
- PAT-A-1056. Mice and Rice (25)
- PAT 1056. Mice and Rice (25)
- PAT 甲级 1056. Mice and Rice (25)
- PAT 1056. Mice and Rice (25)
- 1056. Mice and Rice (25)PAT
- PAT 1056. Mice and Rice
- 【PAT】1056. Mice and Rice
- 【C++】PAT(advanced level)1056. Mice and Rice (25)
- 今天开始学习JAVA!
- 比真机还快的Android模拟器——Genymotion
- Journey - 黄山相关资料
- 关于cocos2d-X 3.x版本使用引擎自带的物理引擎Physics
- 何时不再说"中国就这样"
- PAT 1056. Mice and Rice (25)
- PAT (Advanced) 1083. List Grades (25)
- 基于linux-2.6.38.8内核的SDIO/wifi驱动分析
- 设计模式的分类与详细介绍
- NYOJ 题目24素数距离问题(水题)
- CSS使元素居中的方法
- Android组件通信--Intent
- AMQP协议四
- eclipse aptana插件 —— 【javascript|html|css|jquery|…】自动提示