codeforces 620D Professor GukiZ and Two Arrays

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 const int maxn = 2000 + 50;
  6 
  7 const long long inf = 1e18;
  8 
  9 int n, m;
 10 
 11 long long suma, sumb;
 12 
 13 int a[maxn], b[maxn];
 14 
 15 long long dbl_a[maxn], dbl_b[maxn];
 16 
 17 long long sum_dbl_a[maxn * maxn + 10], sum_dbl_b[maxn * maxn + 10];
 18 
 19 int main()
 20 {
 21     map<int, int> pos_a;
 22     scanf("%d", &n);
 23     for(int i = 1; i <= n; ++i){
 24         scanf("%d", &a[i]);
 25         suma += a[i];
 26         dbl_a[i] = a[i] * 2;
 27         pos_a[dbl_a[i]] = i;
 28     }
 29 
 30     map<long long, pair<int, int>> m_a;
 31     map<int, long long> two_pos_a;
 32     int cnt_a = 0, cnt_b = 0;
 33     for(int i = 1; i <= n; ++i){
 34         for(int j = i + 1; j <= n; ++j){
 35                 m_a[dbl_a[i] + dbl_a[j]] = make_pair(i, j);
 36                 sum_dbl_a[++cnt_a] = dbl_a[i] + dbl_a[j];
 37         }
 38     }
 39     scanf("%d", &m);
 40     for(int i = 1; i <= m; ++i){
 41         scanf("%d", &b[i]);
 42         sumb += b[i];
 43         dbl_b[i] = b[i] * 2;
 44     }
 45 
 46     long long diff = (suma - sumb);
 47 
 48     if(diff == 0) {
 49         printf("0\n0\n");
 50         return 0;
 51     }
 52 
 53     sort(dbl_a+1, dbl_a+1+n);
 54 
 55     long long one_swap_ans = abs(suma - sumb);
 56     long long two_swap_ans = abs(suma - sumb);
 57     int one_swap_a = 0, one_swap_b = 0;
 58     int two_swap_a_1 = 0, two_swap_a_2 = 0, two_swap_b_1 = 0, two_swap_b_2 = 0;
 59 
 60     for(int i = 1; i <= m; ++i){
 61         long long tmp = diff + dbl_b[i];
 62         int pos = lower_bound(dbl_a+1, dbl_a+n+1, tmp) - dbl_a;
 63         //printf("%d\n", pos);
 64         long long big_res = inf;
 65         if(pos <= n) big_res = dbl_a[pos-1];
 66 
 67         long long small_res = inf;
 68         if(pos - 1 >= 1) small_res = dbl_a[pos - 1];
 69 
 70         if(abs(big_res - tmp) > abs(small_res - tmp)){
 71                 //printf("%d %d %d\n",pos_a[10], dbl_a[pos-1], pos);
 72                 long long temp = abs(small_res - tmp);
 73                 if(one_swap_ans > temp){
 74                         one_swap_ans = temp;
 75                         one_swap_a = pos_a[dbl_a[pos-1]], one_swap_b = i;
 76                 }
 77         } else {
 78                 long long temp = abs(big_res - tmp);
 79                 if(one_swap_ans > temp){
 80                         one_swap_ans = temp;
 81                         one_swap_a = pos_a[dbl_a[pos]], one_swap_b = i;
 82                 }
 83         }
 84 
 85     }
 86 
 87 
 88 
 89     sort(sum_dbl_a + 1, sum_dbl_a + cnt_a + 1);
 90 
 91 
 92     for(int i = 1; i <= m; ++i){
 93         for(int j = i + 1; j <= m; ++j){
 94                 long long tmp = diff + dbl_b[i] + dbl_b[j];
 95                 int pos = lower_bound(sum_dbl_a+1, sum_dbl_a+cnt_a+1, tmp) - sum_dbl_a;
 96                 long long big_res = inf;
 97                 if(pos <= cnt_a) big_res = sum_dbl_a[pos];
 98 
 99                 long long small_res = inf;
100                 if(pos - 1 >= 1) small_res = sum_dbl_a[pos - 1];
101 
102                 if(abs(big_res - tmp) > abs(small_res - tmp)){
103                         long long temp = abs(small_res - tmp);
104 
105                         if(two_swap_ans > temp){
106                                 two_swap_ans = temp;
107                                 two_swap_a_1 = m_a[sum_dbl_a[pos-1]].first;
108                                 two_swap_a_2 = m_a[sum_dbl_a[pos-1]].second;
109                                 two_swap_b_1 = i;
110                                 two_swap_b_2 = j;
111                         }
112                 } else {
113                         long long temp = abs(big_res - tmp);
114 
115                         if(two_swap_ans > temp){
116                                 two_swap_ans = temp;
117                                 two_swap_a_1 = m_a[sum_dbl_a[pos]].first;
118                                 two_swap_a_2 = m_a[sum_dbl_a[pos]].second;
119 
120                                 two_swap_b_1 = i;
121                                 two_swap_b_2 = j;
122                         }
123                 }
124 
125                 }
126     }
127 
128 
129         long long ans = min(abs(diff), min(abs(one_swap_ans), abs(two_swap_ans)));
130         printf("%I64d\n", ans);
131         if(ans == diff){
132                 printf("0\n");
133         } else if(one_swap_ans == ans){
134                 printf("1\n");
135                 printf("%d %d\n", one_swap_a, one_swap_b);
136         } else {
137                 printf("2\n");
138                 printf("%d %d\n", two_swap_a_1, two_swap_b_1);
139                 printf("%d %d\n", two_swap_a_2, two_swap_b_2);
140         }
141 
142     return 0;
143 }

 

只交换一次的情况
原来 |Suma - Sumb|
交换后 Suma - xi + yj - (sumb - yj + xi) = suma - sumb + 2(yj - xi) -> 0
则在确定xi的情况下 需要在 b序列里找到 最接近 xi 的 (suma - sumb) /2 + yj的 yj

交换两次的情况
原来 |suma - sumb|
交换后 Suma - xi - xj + yk + yl - (sumb + xi + xj - yk - yl) = suma - sumb + 2((yk + yl) - (xi + xj)) -> 0
则在确定 xi xj的情况下 需要在b序列里找到 最接近 xi + xj的 (suma - sumb)/2 + yk + yl的 yk 和 yl

 

posted @ 2017-08-29 23:59 GeniusYang 阅读(...) 评论(...) 编辑 收藏