题意:先读入球队信息,再按“team_name_1#goals1@goals2#team_name_2”的格式读入球队的比赛情况,然后统计各个球队的比赛得分,进球数,丢球数等等各项指标。最后按题目给定的顺序输出各个球队的信息。
这题一个非常坑的地方就是在排序时队名的排序是按小写排的,看了别人博客上提醒才发现有这回事。。。
还有一个我被坑很久的就是球队的排序不是各指标全部都是从大到小排的。。。这里wa无数次,实在令人羞愧。。。
这里读入球队信息的时候有个小trick,可以用scanf("%d@%d", &goals1, &goals2)这样去读入中间那快数据会更方便(虽然方便不了多少)。。。
代码:
#include <iostream> #include <cstdio> #include <string> #include <algorithm> #include <cctype> using namespace std; struct Table { int tp, g, w, t, l, h, i; string n; }tb[50]; bool cmp(Table a, Table b) { if (a.tp != b.tp) return a.tp > b.tp; if (a.w != b.w) return a.w > b.w; if ((a.h - a.i) != (b.h - b.i)) return (a.h - a.i) > (b.h - b.i); if (a.h != b.h) return a.h > b.h; if (a.g != b.g) return a.g < b.g; string ta, tb; for (int i = 0; i < a.n.size(); i++) ta += toupper(a.n[i]); for (int i = 0; i < b.n.size(); i++) tb += toupper(b.n[i]); return ta < tb; } int main() { int n; scanf("%d", &n); getchar(); while (n--) { string name; int t, g; getline(cin, name); cout << name << endl; scanf("%d\n", &t); for (int i = 0; i < t; i++) { getline(cin, tb[i].n); tb[i].tp = tb[i].g = tb[i].w = tb[i].t = tb[i].l = tb[i].h = tb[i].i = 0; } scanf("%d\n", &g); for (int i = 0; i < g; i++) { string t1, t2; int n1, n2; int s1, s2; char tmp; while ((tmp = getchar()) != '#') t1 += tmp; scanf("%d@%d#", &s1, &s2); getline(cin, t2); for (int i = 0; i < t; i++) if (t1 == tb[i].n) { n1 = i; break; } for (int i = 0; i < t; i++) if (t2 == tb[i].n) { n2 = i; break; } // cout << n1 << ' ' << s1 << ' ' << t1 << endl; // cout << n2 << ' ' << s2 << ' ' << t2 << endl; tb[n1].g++; tb[n2].g++; tb[n1].h += s1; tb[n1].i += s2; tb[n2].h += s2; tb[n2].i += s1; if (s1 > s2) { tb[n1].tp += 3; tb[n1].w++; tb[n2].l++; } else if (s1 < s2) { tb[n2].tp += 3; tb[n2].w++; tb[n1].l++; } else { tb[n1].tp++; tb[n2].tp++; tb[n1].t++; tb[n2].t++; } } sort (tb, tb + t, cmp); for (int i = 0; i < t; i++) { printf("%d) ", i + 1); cout << tb[i].n; printf(" %dp, %dg (%d-%d-%d), %dgd (%d-%d)\n", tb[i].tp, tb[i].g, tb[i].w, tb[i].t, tb[i].l, tb[i].h - tb[i].i, tb[i].h, tb[i].i); } if (n) cout << endl; } return 0; }