2022 RoboCom 世界机器人开发者大赛-本科组（省赛）RC-u4 攻略分队

本文链接：https://blog.youkuaiyun.com/qq_45778406/article/details/125738961

该博客介绍了一道C++编程题，涉及数组、结构体和排序。题目中，通过输入不同部队的属性，计算最优的讨伐组合。程序首先读取部队信息，然后构造结构体并进行排序，最后输出最优的讨伐和防守部队。主要考察了数组操作、条件判断和排序算法的应用。

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在这里插入图片描述

本题思路较为清晰，考场时结构体内部数组开小了，导致wa。

#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int N = 1e5 + 10;
int v[10];
int a[10][4];
struct node
{
    int b[7];
    int num1 = 0;
    int c[7];
    int num2 = 0;
    int dx;
    int zhi;
    int tan; //坦克
    int zg;
    int oy;
    int bian;
    int fz;
    bool operator<(const node &t) const
    {
    	if(t.fz != fz) return fz > t.fz;
    	if(tan != t.tan) return tan > t.tan;
        if(zg != t.zg)
        {
            return zg > t.zg;
        }
        if(zhi != t.zhi)
        {
            return zhi > t.zhi;
        }
        if(dx != t.dx)
        {
            return dx < t.dx;
        }
        if(oy != t.oy) return oy > t.oy;
        for(int i = 1; i <= num1 && i <= t.num1; i ++)
		{
			if(b[i] != t.b[i]) return b[i] < t.b[i];
		}
		return num1 < t.num1;
    }
}q[100];
int main()
{
    for(int i = 1; i <= 6; i ++)
        cin >> v[i];
    int sum1 = 0, sum2 = 0, sum3 = 0;
    for(int i = 1; i <= 6; i ++)
    {
        string s;
        cin >> s;
        for(int j = 1; j <= 3; j ++) a[i][j] = s[j - 1] - '0';
    }
    for(int i = 1; i <= 62; i++)
    {
        int num1 = 0, num2 = 0; // 队伍数
        int s1 = 0, s2 = 0; //  人数之和
        int zhi1 = 0, zhi2 = 0; // 指挥数
        int gon1 = 0, gon2 = 0; // 工兵数
        int tan1 = 0, tan2 = 0; // 坦克数
        int minn = 6; // 最小的编号
        for(int j = 1; j <= 6; j ++)
        {
        	if(v[j] == 0) continue;
            if((i >> (j - 1))&1 == 1) //讨伐ou
            {
            	//cout << pow(2, j - 1) << " ";
                q[i].b[++num1] = j;
                s1 += v[j];
                if(a[j][1] == 1) tan1 = 1;
				if(a[j][2] == 1) gon1 = 1;
				if(a[j][3] == 1) zhi1 = 1;
				minn = min(j, minn);
            }
            else
			{
				q[i].c[++num2] = j;
				s2 += v[j];
				if(a[j][1] == 1) tan2 = 1;
				if(a[j][2] == 1) gon2 = 1;
				if(a[j][3] == 1) zhi2 = 1;
			}
        }
		//cout << endl;
        q[i].num1 = num1;
        q[i].num2 = num2;
        q[i].dx = abs(s1 - s2);
        if(tan1 == tan2 && tan1 == 1) q[i].tan = 1;
        if(gon1 == 1 && gon2 == 1 && zhi1 == 1 && zhi2 == 1) q[i].zg = 1;
        if(zhi1 == 1 && zhi2 == 1) q[i].zhi = 1;
        if(s1 > s2) q[i].oy = 1;
        if(num1 == 0 || num2 == 0) q[i].fz = 0;
        else q[i].fz = 1;
        q[i].bian = minn;
    }
    sort(q + 1, q + 62 + 1);
    if(q[1].fz == 0 || q[1].tan == 0)
	{
		cout << "GG" << endl;
		return 0;
	}
    for(int i = 1; i <= q[1].num1; i++)
	{
		cout << q[1].b[i];
		if(i != q[1].num1) cout << " ";
		else cout << endl;
	}
	for(int i = 1; i <= q[1].num2; i++)
	{
		cout << q[1].c[i];
		if(i != q[1].num2) cout << " ";
		else cout << endl;
	}
    return 0;
}
/*
6 8 7 5 3 0
010
101
010
001
011
000
6 8 7 5 3 0
010
101
110
001
111
000

*/