HUT 排序训练赛 F

Problem's Link

Mean: 

略.

analyse:

使用结构体排序。

首先,定义一个结构体,用来存放输入的数据,然后就是输入,注意:这儿有一个小细节,输入数字,然后紧跟着输入字符串,这时需要一个getchar()语句来接收一个回车符,否则输入完数据后的那个回车符将会被下面的字符串接收导致错误,然后就是结构体排序,还要注意的是题目要求按照字典顺序来排序,这儿就需要用到strcmp函数,最后就是输出的问题,注意前面是三个空格。

Time complexity: O(nlogn*mlogm)

view code

c++:

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct aa {
    char c1[81],c2[81];
    int n;
} a[115];
bool cmp(aa x,aa y) {
    if(strcmp(x.c1,y.c1)!=0)
        return  strcmp(x.c1,y.c1)<0;
    return  strcmp(x.c2,y.c2)<0;
}
int main () {
    int i,t,n,sum;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        getchar();
        for(i=1; i<=n; i++)
            scanf("%s %s %d",a[i].c2,a[i].c1,&a[i].n);
        sort(a+1,a+n+1,cmp);
        strcpy(a[0].c1,"00");
        strcpy(a[0].c2,"00");
        a[0].n=-1;
        strcpy(a[n+1].c1,"00");
        strcpy(a[n+1].c2,"00");
        a[n+1].n=-1;
        sum=a[1].n;
        for(i=1; i<=n; i++) {
            if(strcmp(a[i].c1,a[i-1].c1)!=0) {
                printf("%s
",a[i].c1);
            }
            if(strcmp(a[i].c1,a[i+1].c1)==0&&strcmp(a[i].c2,a[i+1].c2)==0) {
                sum+=a[i+1].n;
            } else {
                printf("   |----%s(%d)
",a[i].c2,sum);
                sum=a[i+1].n;
            }
        }
        if(t) {
            printf("
");
        }
    }
    return 0;
}

java:

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int cas = in.nextInt();
        for (; cas > 0; --cas) {
            TreeMap<String, TreeMap<String, Integer>> originMap = new TreeMap<>();
            int n = in.nextInt();
            for (; n > 0; --n) {
                String name = in.next();
                String origin = in.next();
                int num = in.nextInt();
                TreeMap<String, Integer> nameMap = originMap.get(origin);
                if (nameMap != null) {
                    int total = num;
                    if (nameMap.get(name) != null) {
                        total = nameMap.get(name) + num;
                    }
                    nameMap.put(name, total);
                } else {
                    nameMap = new TreeMap<>();
                    nameMap.put(name, num);
                    originMap.put(origin, nameMap);
                }
            }

            int originMapSize = originMap.size();
            for (Map.Entry<String, TreeMap<String, Integer>> originMapEntry : originMap.entrySet()) {
                --originMapSize;
                System.out.println(originMapEntry.getKey());
                TreeMap<String, Integer> nameMap = originMapEntry.getValue();
                for (Map.Entry<String, Integer> nameMapEntry : nameMap.entrySet()) {
                    System.out.println("   |----" + nameMapEntry.getKey() + "(" + nameMapEntry.getValue() + ")");
                }
            }

            if (cas != 1) {
                System.out.println();
            }
        }
    }
}
原文地址:https://www.cnblogs.com/crazyacking/p/3588644.html