[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。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 1e9;
 const db eps = 1e-;
 const int maxn = 1e5 + ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), last = ' ';
   while(!isdigit(ch)) {last = ch; ch = getchar();}
   while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
   if(last == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) x = -x, putchar('-');
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, k, a[maxn];
 int dif[maxn], pre[maxn], nxt[maxn];
 ll ans = ;

 #define pr pair<int, int>
 #define mp make_pair
 #define fir first
 #define sec second
 priority_queue<pr, vector<pr>, greater<pr> > q;

 int main()
 {
   n = read(); k = read();
   for(int i = ; i <= n; ++i) a[i] = read();
   for(int i = ; i < n; ++i)
     {
       dif[i] = a[i + ] - a[i];
       pre[i] = i - ; nxt[i] = i + ;
       q.push(mp(dif[i], i));
     }
   nxt[n - ] = ;
   for(int i = ; i <= k; ++i)
     {
       pr now = q.top(); q.pop();
       if(now.fir != dif[now.sec]) {k++; continue;}  //合并后就跳过
       ans += now.fir;
       int ls = pre[now.sec], rs = nxt[now.sec];
       nxt[now.sec] = nxt[rs]; pre[nxt[now.sec]] = now.sec;
       pre[now.sec] = pre[ls]; nxt[pre[now.sec]] = now.sec;
       dif[now.sec] = (ls && rs) ? dif[ls] + dif[rs] - dif[now.sec] : INF;
       dif[ls] = dif[rs] = INF;
       q.push(mp(dif[now.sec], now.sec));
     }
   write(ans), enter;
   return ;
 }