[APIO/CTSC 2007]数据备份
标签：小根堆 getch nbsp char tchar 循环 span ref view
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ai，就把ai-1 + ai+1 - ai放到小根堆中，这样如果以后选了ai-1 + ai+1 - ai这个数，就把前面选的ai抵消了，所以这两次操作就相当于选了ai-1和ai+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 2 #include 3 #include 4 #include
5 #include 6 #include 7 #include 8 #include 9 #include10 #include11 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 pair42 #define mp make_pair
43 #define fir first
44 #define sec second
45 priority_queue, greater > q;
46
47 int main()
48 {
49 n = read(); k = read();
50 for(int i = 1; i read();
51 for(int i = 1; i 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 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 }
View Code
[APIO/CTSC 2007]数据备份
标签：小根堆 getch nbsp char tchar 循环 span ref view
原文地址：https://www.cnblogs.com/mrclr/p/9849663.html
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