PAT(A) - 1075. PAT Judge (25)

The ranklist of PAT is generated from the status list, which shows the scores of the submittions. This time you are supposed to generate the ranklist for PAT.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 positive integers, N (<=10⁴), the total number of users, K (<=5), the total number of problems, and M (<=10⁵), the total number of submittions. It is then assumed that the user id's are 5-digit numbers from 00001 to N, and the problem id's are from 1 to K. The next line contains K positive integers p[i] (i=1, ..., K), where p[i] corresponds to the full mark of the i-th problem. Then M lines follow, each gives the information of a submittion in the following format:

user_id problem_id partial_score_obtained

where partial_score_obtained is either -1 if the submittion cannot even pass the compiler, or is an integer in the range [0, p[problem_id]]. All the numbers in a line are separated by a space.

Output Specification:

For each test case, you are supposed to output the ranklist in the following format:

rank user_id total_score s[1] ... s[K]

where rank is calculated according to the total_score, and all the users with the same total_score obtain the same rank; and s[i] is the partial score obtained for the i-th problem. If a user has never submitted a solution for a problem, then "-" must be printed at the corresponding position. If a user has submitted several solutions to solve one problem, then the highest score will be counted.

The ranklist must be printed in non-decreasing order of the ranks. For those who have the same rank, users must be sorted in nonincreasing order according to the number of perfectly solved problems. And if there is still a tie, then they must be printed in increasing order of their id's. For those who has never submitted any solution that can pass the compiler, or has never submitted any solution, they must NOT be shown on the ranklist. It is guaranteed that at least one user can be shown on the ranklist.

Sample Input:

7 4 2020 25 25 3000002 2 1200007 4 1700005 1 1900007 2 2500005 1 2000002 2 200005 1 1500001 1 1800004 3 2500002 2 2500005 3 2200006 4 -100001 2 1800002 1 2000004 1 1500002 4 1800001 3 400001 4 200005 2 -100004 2 0

Sample Output:

1 00002 63 20 25 - 182 00005 42 20 0 22 -2 00007 42 - 25 - 172 00001 42 18 18 4 25 00004 40 15 0 25 -

思路分析： 定义一个学生信息的结构体。包括id、每道题的得分、总分、ac题目的数量、是否打印标记。把逻辑搞清楚就OK了。

#include <cstdio>#include <string.h>#include <algorithm>#define MAX 10001using namespace std;typedef struct Stu {    int id;    // 5位的id,用int表示    int p[6];  // 每道题的得分，初始为-2    int sum;   // 总分    int ac;    // ac题目的数量    bool flag; // 是否打印，初识为false        // 构造函数    Stu() {        id = 0;        p[0] = p[1] = p[2] = p[3] = p[4] = p[5] = -2;        sum = 0;        ac = 0;        flag = false;    }} Stu;Stu stu[MAX];// 排序比较函数int cmp( Stu a, Stu b ) {    if( a.sum != b.sum ) {        return a.sum > b.sum ? 1 : 0;    }    else if( a.ac != b.ac ) {        return a.ac > b.ac ? 1 : 0;    }    else {        return a.id < b.id ? 1 : 0;    }}int main() {    //freopen( "123.txt", "r", stdin );    int n, m, sub;    int problem[6] = { 0 };    scanf( "%d%d%d", &n, &m, &sub );    // 输入每道题的满分    for( int i = 1; i <= m; i++ ) {        scanf( "%d", &problem[i] );    }        // 输入每次提交    int id, pIndex, pScore;    for( int i = 1; i <= sub; i++ ) {        scanf( "%d%d%d", &id, &pIndex, &pScore );        stu[id].id = id;        // 提交的这道题分数比以前大，更新        if( stu[id].p[pIndex] < pScore ) {            stu[id].p[pIndex] = pScore;        }    }    // 统计每个学生的总分、AC数量、是否打印    for( int i = 1; i <= n; i++ ) {        for( int j = 1; j <= m; j++ ) {            if( stu[i].p[j] >= 0 ) {                stu[i].flag = true;                stu[i].sum += stu[i].p[j];            }            if( stu[i].p[j] == problem[j] ) {                stu[i].ac++;            }        }    }    // 排序    sort( stu + 1, stu + n + 1, cmp );    int curRank = 1;    // 当前排名    int preSum = -1;    // 前一个学生的总分    int count = 0;      // 当前有效成绩学生人数    for( int i = 1; i <= n; i++ ) {        if( !stu[i].flag )   // 不打印这个学生的排名            continue;        count++;        if( preSum == -1 ) {            preSum = stu[i].sum;        }        else if( stu[i].sum < preSum ) {            preSum = stu[i].sum;            curRank = count;        }        printf( "%d %05d %d", curRank, stu[i].id, stu[i].sum );        for( int j = 1; j <= m; j++ ) {            if( stu[i].p[j] == -1 ) printf( " 0" );            else if( stu[i].p[j] == -2 ) printf( " -" );            else printf( " %d", stu[i].p[j] );        }        printf( "\n" );    }    return 0;}