HDU 4923 Room and Moor(推理+栈维护)

HDU 4924 Room and Moor

题目链接

题意：给定一个01组成的a序列，要求一个b序列，b序列每个数值为[0, 1]之间的数，并且b序列为非递减序列，要求∑(ai−bi)2最小，求这个最小值

思路：推理，很容易看出，开头一段的0和末尾一段的1等于没有，然后中间每段类似111000这样1在前，0在后的序列，都可以列出一个公式，很容易推出选择的x为共同的一个值，为1的个数/（1的个数+0的个数）a,那么问题就变成要维护一个递增的x，利用一个栈去做维护，如果遇到一个位置递减了，那么就把它和之前的段进行合并，维护栈中递增，最后把栈中元素都拿出来算一遍就是答案了

代码：

#include <cstdio>#include <cstring>#include <stack>using namespace std;const int N = 100005;const double eps = 1e-8;int t, n, a[N];struct Seg {    double one, zero, x;    Seg() {}    Seg(double one, double zero, double x) {    this->one = one;    this->zero = zero;    this->x = x;    }} s[N];double cal(double one, double zero) {    return one / (one + zero);}double solve(int n) {    int sn = 0;    int one = 0, zero = 0, now = 0;    while (now < n && !a[now]) {now++;}    while (n >= 0 && a[n]) n--;    if (now > n) return 0.0;    while (now <= n) {    double one = 0, zero = 0;    while (a[now]) {        one += 1;        now += 1;    }    while (now <= n && !a[now]) {        zero += 1;        now += 1;    }    s[sn].one = one;    s[sn].zero = zero;    s[sn++].x = cal(one, zero);    }    stack<Seg> st;    st.push(s[0]);    for (int i = 1; i < sn; i++) {        Seg now = s[i];    while (!st.empty() && st.top().x - now.x > -eps) {        Seg pre = st.top();        st.pop();        pre.one += now.one;        pre.zero += now.zero;        pre.x = cal(pre.one, pre.zero);        now = pre;    }    st.push(now);    }    double ans = 0;    while (!st.empty()) {    Seg now = st.top();    st.pop();    ans += (1 - now.x) * (1 - now.x) * now.one + now.x * now.x * now.zero;    }    return ans;}int main() {    scanf("%d", &t);    while (t--) {    scanf("%d", &n);    for (int i = 0; i < n; i++)        scanf("%d", &a[i]);    printf("%.6lf\n", solve(n - 1));    }    return 0;}