Problem Description As is known to all, long long ago sailormoon once was an association of fighters. Till now, sailormoon is also an association of girls. Owe to some unknown reasons, girls are necessary to fight for peace. Their boss, lcy, wants to strengthen their ability, so he give them his precious collections---weapons for many years. Because these collections are really age-old, it is hard to recognize from one to another. So girls intend to sort them before they use. Each weapon has its name, origin and level of harmfulness ( level contains three ranks: wonderful, good, so-so). In order to make it clear, girls want to sort like this: firstly,sort according to the origin (sort by lexicographic order), if two or more have the same origin, they will be sorted together; secondly, sort according ranks, wonderful is the best, good is next, the third is so-so; thirdly, if two or more have same origin and rank, sort them according to the lexicographic order. Input Input contains multiply cases. Each case contains several lines. First line is an integer N(0<N<=500), representing the number of weapons. Then N lines follows. Each line represent a kind of weapon, and contains a set of strings representing name, origin and level of harmfulness. Each string will not exceed 20 characters. Sure that same origin will not exist the same weapon. Output Please output your list after sorting (format according to sample, pay attention to the spaces,ten spaces need ^ ^). Sample Input 5 knife qizhou so-so gun qizhou wonderful knife zhengzhou good stick zhengzhou good rope shengzhou so-so Sample Output Case 1 qizhou: gun wonderful knife so-so shengzhou: rope so-so zhengzhou: knife good stick good
#include<stdio.h> #include<iostream> #include<cstring> #include<algorithm> #define N 10000 using namespace std; struct SS{ char place[200]; char name[200]; char rank[200]; //string rank; int rankle; int num; }f[N]; int cmp(SS a,SS b) { if(strcmp(a.place,b.place)!=0) return strcmp(a.place,b.place)<0; if(a.rankle!=b.rankle) return a.rankle>b.rankle; if(strcmp(a.name,b.name)!=0) return strcmp(a.name,b.name)<0; } int main() { //freopen("1.txt","r",stdin); int test,i,n,m,flag,count=1; while(cin>>test) { for(i=0;i<test;i++) { cin>>f[i].name>>f[i].place>>f[i].rank; if(strcmp(f[i].rank,"wonderful")==0) f[i].rankle=3; //strcpy(f[i].rank,'3');这个地方也不能直接将3直接给rank也不能把‘3’给rank，所以用一个int型的rankle, if(strcmp(f[i].rank,"good")==0) f[i].rankle=2; //strcpy(f[i].rank,'2');直接对rankle进行比较 if(strcmp(f[i].rank,"so-so")==0) f[i].rankle=1; //strcpy(f[i].rank,'1'); } sort(f,f+test,cmp); flag=1; //count=1; cout<<"Case "<<count<<endl; char di[200]; char names[200]; strcpy(di,f[0].place); strcpy(names,f[0].name); for(i=0;i<test;i++) { if(strcmp(f[i].place,di)!=0) { strcpy(di,f[i].place); flag=1; i--; } if(strcmp(f[i].place,di)==0) { if(flag==1) { cout<<f[i].place<<":"<<endl; flag=0; } cout<<" "<<f[i].name<<" "<<f[i].rank<<endl; } } count++; } return 0; }