[APIO/CTSC 2007]数据备份

嘟嘟嘟

这竟然是一道贪心题，然而我在不看题解之前一直以为是dp。

首先最优的配对一定是相邻两个建筑物配对，所以我们求出差分数组，就变成了在n - 1个数中选出不相邻的k个数，使这k个数的和最小。

贪心是在回事呢？首先把所有点放在一个小根堆中，然后如果取出一个点a_i，就把a_i-1 + a_i+1 - a_i放到小根堆中，这样如果以后选了a_i-1 + a_i+1 - a_i这个数，就把前面选的a_i抵消了，所以这两次操作就相当于选了a_i-1和a_i+1这两个数。

每选一次就合并了两个数，那么进行k次就选了k个数，所以循环k次后，累加的答案就是最优解。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(‘ ‘)
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 1e9;
19 const db eps = 1e-8;
20 const int maxn = 1e5 + 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) x = -x, putchar(‘-‘);
33   if(x >= 10) write(x / 10);
34   putchar(x % 10 + ‘0‘);
35 }
36
37 int n, k, a[maxn];
38 int dif[maxn], pre[maxn], nxt[maxn];
39 ll ans = 0;
40
41 #define pr pair<int, int>
42 #define mp make_pair
43 #define fir first
44 #define sec second
45 priority_queue<pr, vector<pr>, greater<pr> > q;
46
47 int main()
48 {
49   n = read(); k = read();
50   for(int i = 1; i <= n; ++i) a[i] = read();
51   for(int i = 1; i < n; ++i)
52     {
53       dif[i] = a[i + 1] - a[i];
54       pre[i] = i - 1; nxt[i] = i + 1;
55       q.push(mp(dif[i], i));
56     }
57   nxt[n - 1] = 0;
58   for(int i = 1; i <= k; ++i)
59     {
60       pr now = q.top(); q.pop();
61       if(now.fir != dif[now.sec]) {k++; continue;}  //合并后就跳过
62       ans += now.fir;
63       int ls = pre[now.sec], rs = nxt[now.sec];
64       nxt[now.sec] = nxt[rs]; pre[nxt[now.sec]] = now.sec;
65       pre[now.sec] = pre[ls]; nxt[pre[now.sec]] = now.sec;
66       dif[now.sec] = (ls && rs) ? dif[ls] + dif[rs] - dif[now.sec] : INF;
67       dif[ls] = dif[rs] = INF;
68       q.push(mp(dif[now.sec], now.sec));
69     }
70   write(ans), enter;
71   return 0;
72 }

原文地址：https://www.cnblogs.com/mrclr/p/9849663.html