Bestcoder4——Happy Three Friends(二叉堆）

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Happy Three Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Dong-hao , Grandpa Shawn , Beautful-leg Mzry are good friends. One day , they want to play a game.There are 6 numbers on the table. Firstly , Dong-hao can change the order of 6 numbers.Secondly , Grandpa Shawn take the first one and the last one , sum them up as his scores.Thirdly , Beautiful-leg Mzry take any of 3 numbers from the last 4 numbers , and sum them up as his scores.Finally , if Grandpa Shawn's score is larger than Beautiful-leg Mzry's , Granpa Shawn wins! If Grandpa Shawn's score is smaller than Beautiful-leg Mzry's , Granpa Shawn loses.If the scores are equal , there is a tie.Nowadays , it's really sad that Grandpa Shawn loses his love. So Dong-hao wants him to win(not even tie). You have to tell Dong-hao whether he can achieve his goal.

 

Input

There is a number T shows there are T test cases below. ( T <= 50)For each test case , there are 6 numbers Ai ( 1 <= Ai <= 100 ).

 

If Dong-hao can achieve his goal , output "Grandpa Shawn is the Winner!"If he can not , output "What a sad story!"

 

Sample Input

31 2 3 3 2 22 2 2 2 2 21 2 2 2 3 4

 

Sample Output

What a sad story!What a sad story!Grandpa Shawn is the Winner!
Hint
For the first test case , {3 , 1 , 2 , 2 , 2 , 3} Grandpa Shawn can take 6 at most . But Beautiful-leg Mzry can take 6 too. So there is a tie.For the second test cases , Grandpa Shawn loses.For the last one , Dong-hao can arrange the numbers as {3 , 2 , 2 , 2 , 1 , 4} , Grandpa Shawn can take 7 , but Beautiful-leg Mzry can take 6 at most. So Grandpa Shawn Wins!

AC代码：之前没有清空队列，已更改。

Output

#include <iostream>#include <stdio.h>#include <string.h>#include <vector>#include <map>#include <algorithm>#include <queue>#include <cmath>#define INF 0xffffff#define lln long long#ifdef ONLINE_JUDGE#define FOI(file) 0#define FOW(file) 0#else#define FOI(file) freopen(file,"r",stdin);#define FOW(file) freopen(file,"w",stdout);#endifusing namespace std;struct Node{    int n;    bool operator < (const Node & a) const    {        return n < a.n;    }};int n;int main(){    //FOI("input");    //FOW("output");    //write your programme here        int i, j, k;    int n;    int t;    int sum, sum2;    Node temp;    priority_queue <Node> q;    scanf("%d", &t);    while(t--)    {        for(i = 0; i < 6; i++)        {            scanf("%d", &temp.n);            q.push(temp);        }                sum = 0;        temp = q.top();        sum += temp.n;        q.pop();        temp = q.top();        sum += temp.n;        q.pop();        //cout << sum << endl;                sum2 = 0;        for(i = 0; i < 3; i++)        {            temp = q.top();            sum2 += temp.n;            q.pop();        }        //cout << sum2 << endl;        while(!q.empty())           q.pop();        if(sum > sum2)            puts("Grandpa Shawn is the Winner!");        else            puts("What a sad story!");    }    return 0;}