[JLOI2011]不重复数字
此题看起来不难。
首先我想到的是用map，然而只能的70分，剩下的超时了。然后我就想到了用去重函数unique，这样就过了。
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #include<cstdlib> 7 #include<stack> 8 #include<queue> 9 #include<vector> 10 #include<cctype> 11 #include<map> 12 using namespace std; 13 #define enter puts("") 14 #define space putchar(' ') 15 #define Mem(a) memset(a, 0, sizeof(a)) 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 5e4 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) putchar('-'), x = -x; 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 struct Node 38 { 39 int x, id; 40 bool operator < (const Node& other)const 41 { 42 return id < other.id || (id == other.id && x < other.x); 43 } 44 bool operator == (const Node& other)const //unique要用的 45 { 46 return x == other.x; 47 } 48 }a[maxn]; 49 50 bool cmp(Node a, Node b) 51 { 52 return a.x < b.x || (a.x == b.x && a.id < b.id); 53 } 54 55 int main() 56 { 57 int T = read(); 58 while(T--) 59 { 60 int n = read(); 61 for(int i = 1; i <= n; ++i) a[i].x = read(), a[i].id = i; 62 sort(a + 1, a + n + 1, cmp); //按cmp排序 63 int _n = unique(a + 1, a + n + 1) - a - 1; 64 sort(a + 1, a + _n + 1); //按结构体自己的排序 65 for(int i = 1; i <= _n; ++i) write(a[i].x), space; enter; 66 } 67 return 0; 68 }
