PAT 天梯赛 L2-028. 秀恩爱分得快 【数据处理】

题目链接

https://www.patest.cn/contests/gplt/L2-028

思路

0.只处理被询问的情侣的亲密度，否则会超时

1.要注意输入数字要用字符串，还要标记性别 因为 输出-0 得到的数字是0

也就是说用int 型输入 是没有办法 辨别编号0的性别的

2.要注意被询问的情侣可能没有出现在照片当中。

输出的时候也要注意负号

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const double PI  = acos(-1);
const double E   = exp(1);
const double eps = 1e-6;

const int INF  = 0x3f3f3f3f;
const int maxn = 1e3 + 5;
const int MOD  = 1e9 + 7;

double ans[2][maxn];

map <int, int> M;

int tran(char s[])
{
    int len = strlen(s);
    int i;
    if (s[0] == '-')
        i = 1;
    else
        i = 0;
    int ans = 0;
    for ( ; i < len; i++)
    {
        ans = ans * 10 + s[i] - '0';
    }
    if (s[0] == '-')
        M[ans] = -1;
    else
        M[ans] = 1;
    return ans;
}

void print(int x, int y)
{
    int a = abs(x);
    int b = abs(y);
    if (M[a] == -1)
        printf("-%d ", a);
    else
        printf("%d ", a);
    if (M[b] == -1)
        printf("-%d\n", b);
    else
        printf("%d\n", b);
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    vector <int> G[maxn];
    int num, tot;
    char s[10];
    for (int i = 0; i < m; i++)
    {
        scanf("%d", &tot);
        for (int j = 0; j < tot; j++)
        {
            scanf("%s", s);
            num = tran(s);
            G[i].pb(num);
        }
    }
    int A, B;
    CLR(ans);
    scanf("%s", s);
    A = tran(s);
    scanf("%s", s);
    B = tran(s);
    vector <int>::iterator it;
    double Max[2] = { 0.0, 0.0};
    for (int i = 0; i < m; i++)
    {
        int flag[2] = { 0, 0 };
        for (it = G[i].begin(); it != G[i].end(); it++)
        {
            if (*it == abs(A))
                flag[0] = 1;
            if (*it == abs(B))
                flag[1] = 1;
            if (flag[0] && flag[1])
                break;
        }
        if (flag[0])
        {
            double k = 1.0 / G[i].size();
            for (it = G[i].begin(); it != G[i].end(); it++)
            {
                if (*it != abs(A) && M[*it] * M[abs(A)] == -1)
                {
                    ans[0][*it] += k;
                }
                Max[0] = max(Max[0], ans[0][*it]);
            }
        }
        if (flag[1])
        {
            double k = 1.0 / G[i].size();
            for (it = G[i].begin(); it != G[i].end(); it++)
            {
                if (*it != abs(B) && M[*it] * M[abs(B)] == -1)
                {
                    ans[1][*it] += k;
                }
                Max[1] = max(Max[1], ans[1][*it]);
            }
        }
    }
    if (ans[0][abs(B)] == Max[0] && ans[1][abs(A)] == Max[1])
    {
        print(A, B);
    }
    else
    {
        for (int i = 0; i < 2; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (ans[i][j] == Max[i] && M[abs(i? B:A)] * M[j] == -1)
                {
                    print(i? B:A, j);
                }
            }
        }
    }
}