PAT-1056. Mice and Rice (25)

1056. Mice and Rice (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 N_P programmers. Then every N_G 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 N_G 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: N_P and N_G (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than N_G 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 N_P distinct non-negative numbers W_i (i=0,...N_P-1) where each W_i is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,...N_P-1 (assume that the programmers are numbered from 0 to N_P-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 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

---

 

首先，这道题需要读懂题。

同样的代码在pat和牛客网运行结果不同，发现牛客网的时限是1秒，而pat是100ms，所以pat运行超时是正常的；在pat上尽量不要使用vector，经常遇到超时的情况。

这是一道模拟竞赛题，使用vector的理念非常简单，就是将胜出的都放入一个新的vector里，然后对新的vector进行同样的模拟规则。

如果不用vector，就要好好想想怎么做了，这道题关键不在谁胜出了，而是谁没有胜出，对于没有胜出的给予名次，做好标记，不再查询。

这里附上了两种实现方法。

对于使用while循环，我们尽量将跳出循环的条件判断放入while循环中，这样更一目了然，也不容易犯错误。

#include <bits/stdc++.h>

using namespace std;

int P, G;
int weight[1003];
vector<int> mouseVec, mouseVec2;
int ranks[10003];

int main()
{
    cin>>P>>G;
    for(int i = 0; i < P; i++) {
        scanf("%d", &weight[i]);
    }
    int x;
    for(int i = 0; i < P; i++) {
        scanf("%d", &x);
        mouseVec.push_back(x);
    }
    while(1) {
        int k, mouseMax;
        for(int i = 0; i < mouseVec.size(); i++) {
            if(i%G == 0) {
                mouseMax = weight[mouseVec[i]];
                k = mouseVec[i];
            }
            else {
                if(mouseMax < weight[mouseVec[i]]) {
                    ranks[k] = -1;
                    mouseMax = weight[mouseVec[i]];
                    k = mouseVec[i];
                }
                else {
                    ranks[mouseVec[i]] = -1;
                }
                if(i%G == G-1) {
                    mouseVec2.push_back(k);
                }
            }
            if(i == mouseVec.size()-1 && i%G != G-1) {
                mouseVec2.push_back(k);
            }
        }
        int r = mouseVec2.size();
        //cout<< "R:"<< r<< endl;
        for(int i = 0; i < mouseVec.size(); i++) {
            if(ranks[mouseVec[i]] == -1) {
                ranks[mouseVec[i]] = r+1;
                //cout<< mouseVec[i]<< " ";
            }
        }
        //cout<< endl;
        if(r == 1) {
            ranks[mouseVec2[0]] = 1;
            break;
        }
        mouseVec.clear();
        for(int i = 0; i < mouseVec2.size(); i++) {
            mouseVec.push_back(mouseVec2[i]);
        }
        mouseVec2.clear();
    }
    for(int i = 0; i < P; i++) {
        if(i != 0) printf(" ");
        printf("%d", ranks[i]);
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int P, G;
int weight[1003];
//vector<int> mouseVec, mouseVec2;
int ranks[1003];
int order[1003];

int main()
{
    cin>>P>>G;
    for(int i = 0; i < P; i++) {
        scanf("%d", &weight[i]);
    }
    int x;
    for(int i = 0; i < P; i++) {
        scanf("%d", &order[i]);
    }
    int c = 0;
    while(1) {
        int ran = (P-c)/G+1;
        if((P-c)%G != 0) {
            ran++;
        }
        int flag = 0;
        int weightMax;
        int cnt = 0;
        int k = 0;
        for(int i = 0; i < P; i++) {
            if(ranks[order[i]] == 0) {
                flag = 1;
                if(cnt % G == 0) {
                    weightMax = weight[order[i]];
                    k = order[i];
                }
                else {
                    if(weightMax < weight[order[i]]) {
                        weightMax = weight[order[i]];
                        ranks[k] = ran;
                        k = order[i];
                    }
                    else {
                        ranks[order[i]] = ran;
                    }
                    c++;
                }
                cnt++;
            }
        }
        if(ran == 2) {
            for(int i = 0; i < P; i++) {
                if(ranks[order[i]] == 0) {
                    ranks[order[i]] = 1;
                    break;
                }
            }
        }
        if(flag == 0) break;
    }
    for(int i = 0; i < P; i++) {
        if(i != 0) printf(" ");
        printf("%d", ranks[i]);
    }
    return 0;
}