H ICPC Standings

链接：https://ac.nowcoder.com/acm/contest/18462/H
来源：牛客网

题目描述

ICPC Regional contest results are determined by the final team standings after a regional contest. A regional contest consists of some number of problems that are to be solved by some number of teams.

Teams are ranked according to the most problems solved. Teams who solve the same number of problems are ranked by least total time. The total time is the sum of the time consumed for each problem solved. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submittal of the first accepted run plus 20 penalty minutes for every previously rejected run for that problem. There is no time consumed for a problem that is not solved. In the event of ties, the team with the smaller time consumed of the last correctly submitted solution ranks higher. That process is repeated as needed (2nd to last correctly submitted problem, 3rd to last correctly submitted problem, etc.) If there is still a tie after all tie-breakers have been exhausted, the teams are ranked equally and displayed in team number order. For example, if there are 3 teams in a contest, and teams 1 and 3 both rank 1, then team 2 would be rank 3 (there is no rank 2 in this case).

For this problem you will write a program to print the final standings of a contest based on the supplied input.

输入描述:

The first line of input contains four space separated integers that define the contest parameters: NT NP NS NR for number of teams, number of problems, number of submissions and number ofhighest rank to display respectively. (2 <= NT <= 100), (1 <= NP <= 20), (1 <= NS <= 10000), (1 <= NR <= NT). Note that it is possible that the highest ranked team(s) actually displayed could be lessthan NR.

The next NS lines each contain four space separated integers the describe submissions. Eachsubmission line has T P t D for team number, problem number, time submitted and dispositionrespectively. The time submitted is the number of minutes since the start of the contest. (1 <= T <= NT), (1 <= P <= NP), (0 <= t < 300), D = 0 or 1 as to whether the submission rejected (wrong) oraccepted (correct) respectively. The value of t will never be less than the previous line’s value for t.Any submissions with t >= 300 should be ignored.

输出描述:

The output will consist of the ranked standings (best to worst) for the contest showing all teams withrank 1 through NR . Each line will contain 16 characters grouped in four columns. The first column isthe rank left justified in a four-character field. The second column is the team number left justified ina four-character field. The third column is the number of problems solved in a right justified threecharacter field. The fourth column is the total time right justified in a five-character field.

示例1

输入

复制

50 12 45 2
16 1 2 1
50 1 5 1
3 1 5 1
16 11 8 0
16 7 10 1
3 7 11 1
50 7 11 1
16 8 14 1
3 11 16 0
3 9 24 1
50 8 27 1
16 11 29 0
50 11 39 1
16 9 41 1
3 8 42 1
50 9 50 1
3 11 52 1
3 10 56 1
16 5 62 1
16 11 72 1
3 3 75 1
50 3 77 1
16 3 103 1
3 3 132 0
3 5 138 1
16 2 147 0
16 2 155 0
16 10 169 1
16 2 188 0
16 2 197 1
50 5 232 1
3 4 253 1
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
50 10 270 0
3 6 299 1
50 10 299 1

输出

复制

1   3    10  975
2   16    9  770

备注:

因为可能出现rank并列的情况，最后输出的是rank<=NR的队伍。例如rank为：1 2 3 4 4 4 7 8 9，如果NR为4，则输出1 2 3 4 4 4。

#include <time.h>
#include <cstdio>
#include <iostream>
#include <set>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <string>
#include <iomanip>
using namespace std;const int maxn = 1e5 + 9;struct info {long long fashi = 0, tishu = 0, id, rank = 0, last = 0;long long timeofti[22], fashiofti[22];bool pass[22];
}a[111];bool cmp(info x, info y) {if (x.tishu != y.tishu)return x.tishu > y.tishu;if (x.fashi != y.fashi)return x.fashi < y.fashi;if (x.last != y.last)return x.last < y.last;return x.id < y.id;
}int main() {for (int i = 0; i < 111; i++) {a[i].id = i;}int nt, np, ns, nr;cin >> nt >> np >> ns >> nr;while (ns--) {int T, p, t, d;cin >> T >> p >> t >> d;if (a[T].pass[p] == 1)continue;if (t < 300) {if (d == 1) {a[T].pass[p] = 1;a[T].tishu++;a[T].fashi += t + a[T].fashiofti[p];a[T].last = max(t + a[T].fashiofti[p], a[T].last);}else {a[T].fashiofti[p] += 20;}}}sort(a + 1, a + nt + 1, cmp);int r = 2;a[1].rank = 1;for (int i = 2; i <= nt; i++) {if (a[i].tishu != a[i - 1].tishu || a[i].fashi != a[i - 1].fashi || a[i].last != a[i - 1].last) {a[i].rank = i;}else {a[i].rank = a[i - 1].rank;}}for (int i = 1; i <= nt; i++) {if (a[i].rank <= nr)cout << left << setw(4) << a[i].rank << left << setw(4) << a[i].id << right << setw(3) << a[i].tishu << right << setw(5) << a[i].fashi << endl;else break;}return 0;
}

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H ICPC Standings

题目描述

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