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); } } } } }转载于:https://www.cnblogs.com/Dup4/p/9433149.html
