垃圾佬的旅游III（Hash + 暴力）

题目链接：http://120.78.128.11/Problem.jsp?pid=3445

最开始的思路就是直接暴力求解，先把所有的数值两两存入结构体，再从小到大枚举。用二分的思路去判断数值以及出现，结果TLE，但优化一下应该也能过，因为题目说只有两组数据。代码如下：

  1 #include <iostream>
  2 #include <string>
  3 #include <cstdio>
  4 #include <cstdlib>
  5 #include <sstream>
  6 #include <iomanip>
  7 #include <map>
  8 #include <stack>
  9 #include <deque>
 10 #include <queue>
 11 #include <vector>
 12 #include <set>
 13 #include <list>
 14 #include <cstring>
 15 #include <cctype>
 16 #include <algorithm>
 17 #include <iterator>
 18 #include <cmath>
 19 #include <bitset>
 20 #include <ctime>
 21 #include <fstream>
 22 #include <limits.h>
 23 #include <numeric>
 24 
 25 using namespace std;
 26 
 27 #define F first
 28 #define S second
 29 #define mian main
 30 #define ture true
 31 
 32 #define MAXN 1000000+5
 33 #define MOD 1000000007
 34 #define PI (acos(-1.0))
 35 #define EPS 1e-6
 36 #define MMT(s) memset(s, 0, sizeof s)
 37 typedef unsigned long long ull;
 38 typedef long long ll;
 39 typedef double db;
 40 typedef long double ldb;
 41 typedef stringstream sstm;
 42 const int INF = 0x3f3f3f3f;
 43 
 44 int n;
 45 int ans,tot;
 46 int a[1010];
 47 struct node{
 48     int x,a,b;
 49     bool operator < (const node &b) const{
 50         return x < b.x;
 51     }
 52 }q[MAXN];
 53 
 54 int find_f(int x,int a,int b){
 55     int l(0),r = tot;
 56     while(l<r){
 57         int mid = (l + r) >> 1;
 58         if(q[mid].x < x)
 59             l = mid+1;
 60         else
 61             r = mid;
 62     }
 63     while(l < tot && q[l].x == x){
 64         if(q[l].a != a && q[l].b != a && q[l].a != b && q[l].b != b)
 65             break;
 66         l++;
 67     }
 68     if(q[l].x != x)
 69         return 0;
 70     return 1;
 71 }
 72 
 73 int main(){
 74     ios_base::sync_with_stdio(false);
 75     cout.tie(0);
 76     cin.tie(0);
 77     while(scanf("%d",&n)!=EOF){
 78         if(!n)
 79             break;
 80         for(int i = 1; i <= n; i++)
 81             scanf("%d",&a[i]);
 82         ans = -INF;
 83         tot = 0;
 84         MMT(q);
 85         for(int i = 1; i < n; i++){
 86             for (int j = i+1; j <= n; j++){
 87                 q[tot].x = a[i] + a[j];
 88                 q[tot].a = i;
 89                 q[tot++].b = j;
 90             }
 91         }
 92         sort(q,q+tot);
 93         for(int i = 1; i <= n; i++){
 94             for(int j = 1; j <= n; j++){
 95                 if(i != j){
 96                     int t = a[i]-a[j];
 97                     if(find_f(t,i,j)){
 98                         if(a[i] > ans)
 99                             ans = a[i];
100                     }
101                 }
102             }
103         }
104         if(ans == -INF)
105             printf("No Solution\n");
106         else
107             printf("%d\n",ans);
108     }
109     return 0;
110 }

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