1049: [HAOI2006]数字序列
Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 1842 Solved: 797
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Description
现在我们有一个长度为n的整数序列A。但是它太不好看了,于是我们希望把它变成一个单调严格上升的序列。
但是不希望改变过多的数,也不希望改变的幅度太大。
Input
第一行包含一个数n,接下来n个整数按顺序描述每一项的键值。n<=35000,保证所有数列是随机的
Output
第一行一个整数表示最少需要改变多少个数。 第二行一个整数,表示在改变的数最少的情况下,每个数改变
的绝对值之和的最小值。
Sample Input
4
5 2 3 5
5 2 3 5
Sample Output
1
4
4
HINT
Source
神。。神题。。。。
首先第一问就很显然,设b[i] = a[i] - i,答案就为b数组的n - LIS
然后第二问就很难(看了ydc的sol才懂)
考虑LIS的f数组的转移,每个最优转移用边给连起来
因为a的单调递增意味着b的单调不降,所以可以考虑设g[i]为答案
则g[i] = min{g[j],w(j,i)} (1 <= j < i)
w(j,i)为使[j,i]区间合法的最小代价
然后有一个结论:w(j,i)的最优方案,一定是存在一个k,使得[j,k]为b[j],[k + 1,i]为b[i]
证明详见ydc的sol
代码:
#include<cstdio> #include<vector> #include<queue> #include<ctime> #include<algorithm> #include<cstdlib> #include<stack> #include<cstring> #include<cmath> using namespace std; typedef long long LL; const LL INF = 10000000000000000LL; const int maxn = 100010; vector<int> p[maxn]; int n; LL a[maxn],b[maxn],f[maxn],g[maxn],sum[maxn]; int lc[maxn * 4],rc[maxn * 4],rt[maxn]; LL data[maxn],ha[maxn],N; LL maxx[maxn]; inline LL getint() { LL ret = 0,f = 1; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); } while (c >= '0' && c <= '9') ret = ret * 10 + c - '0',c = getchar(); return ret * f; } inline LL w(int j,int i) { LL now = 0,ret; for (int t = j + 1; t < i; t++) now += abs(ha[b[t]] - ha[b[i]]); ret = now; for (int t = j + 1; t < i; t++) now += abs(ha[b[t]] - ha[b[j]]) - abs(ha[b[t]] - ha[b[i]]) , ret = min(now,ret); return ret; } int main() { n = getint(); for (int i = 1; i <= n; i++) a[i] = getint(); for (int i = 1; i <= n; i++) data[i] = b[i] = a[i] - i; sort(data + 1,data + n + 1); for (int i = 1; i <= n; i++) if (i == 1 || data[i - 1] != data[i]) ha[++N] = data[i]; ha[0] = ha[1]; for (int i = 1; i <= n; i++) b[i] = lower_bound(ha + 1,ha + N + 1,b[i]) - ha; LL ans = 0; for (int i = 1; i <= n; i++) { for (int j = b[i]; j; j -= j & -j) f[i] = max(f[i],maxx[j] + 1); for (int j = b[i]; j <= N; j += j & -j) maxx[j] = max(maxx[j],f[i]); ans = max(f[i],ans); } printf("%d\n",n - ans); p[0].push_back(0); for (int i = 1; i <= n; i++) { g[i] = INF; p[f[i]].push_back(i); for (int k = 0; k < p[f[i] - 1].size(); k++) { int j = p[f[i] - 1][k]; if (b[j] > b[i]) continue; g[i] = min(g[i],g[j] + w(j,i)); } } LL Ans = INF; for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + ha[b[i]]; for (int i = 0; i <= n; i++) { if (f[i] != ans) continue; Ans = min(g[i] + ha[b[i]] * (n - i) - (sum[n] - sum[i]),Ans); } printf("%lld\n",Ans); return 0; }