PAT甲级 1062. Talent and Virtue (25)

题目：

About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage（圣人）"; being less excellent but with one's virtue outweighs talent can be called a "nobleman（君子）"; being good in neither is a "fool man（愚人）"; yet a fool man is better than a "small man（小人）" who prefers talent than virtue.

Now given the grades of talent and virtue of a group of people, you are supposed to rank them according to Sima Guang's theory.

Input Specification:

Each input file contains one test case. Each case first gives 3 positive integers in a line: N (<=10⁵), the total number of people to be ranked; L (>=60), the lower bound of the qualified grades -- that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (<100), the higher line of qualification -- that is, those with both grades not below this line are considered as the "sages", and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as the "noblemen", and are also ranked in non-increasing order according to their total grades, but they are listed after the "sages". Those with both grades below H, but with virtue not lower than talent are considered as the "fool men". They are ranked in the same way but after the "noblemen". The rest of people whose grades both pass the L line are ranked after the "fool men".

Then N lines follow, each gives the information of a person in the format:

ID_Number Virtue_Grade Talent_Grade

where ID_Number is an 8-digit number, and both grades are integers in [0, 100]. All the numbers are separated by a space.

Output Specification:

The first line of output must give M (<=N), the total number of people that are actually ranked. Then M lines follow, each gives the information of a person in the same format as the input, according to the ranking rules. If there is a tie of the total grade, they must be ranked with respect to their virtue grades in non-increasing order. If there is still a tie, then output in increasing order of their ID's.

思路：

这道题重点在排序上，题目里有些没说明白。在排序上，先根据不同人的种类进行划分，在根据总分进行排序，总分相同根据virtue grade 进行排序，再相同的话就根据id进行排序。这里注意id上是数组里的字符进行比较，而不是数组指针。另外，要用scanf和printf。

代码：

#include<iostream>#include<vector>#include<string>#include<algorithm>using namespace std;struct person{char id[9];int virtue_grade;int talent_grade;int rank;bool operator <(const person &p)const{if (rank != p.rank){return rank < p.rank;}else{if (virtue_grade + talent_grade != p.virtue_grade + p.talent_grade){return virtue_grade + talent_grade >= p.virtue_grade + p.talent_grade;}else{if (virtue_grade != p.virtue_grade){return virtue_grade > p.virtue_grade;}else{int i = 0;for (; i < 8; ++i){if (id[i] != p.id[i]){return id[i] < p.id[i];}}return 1;}}}}};int main(){//ifstream cin;//cin.open("case1.txt");int N, L, H;cin >> N >> L >> H;int i,t;vector<person> P(N);t = N;for (i = 0; i < N; ++i){//scanf("%s %d %d", P[i].id, &P[i].virtue_grade, &P[i].talent_grade);cin >> P[i].id >> P[i].virtue_grade >> P[i].talent_grade;if (P[i].talent_grade < L || P[i].virtue_grade < L){P.pop_back();--i;--N;}else{if (P[i].virtue_grade >= H){if (P[i].talent_grade >= H){P[i].rank = 0; //sages}else{P[i].rank = 1;//nobleman}}else{if (P[i].talent_grade < H && P[i].virtue_grade>= P[i].talent_grade){P[i].rank = 2;//fool man}else{P[i].rank = 3;//rest}}}}P.resize(N);sort(P.begin(),P.end());printf("%d\n",N);for (i = 0; i < N; ++i){printf("%s %d %d\n",P[i].id,P[i].virtue_grade,P[i].talent_grade);}system("pause");return 0;}