[NOIP 2016] 蚯蚓

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=4721

[算法]

首先，我们可以维护一个堆，堆中存放蚯蚓的长度，由于除当前蚯蚓其他的蚯蚓长度都要增加q，我们不妨将当前蚯蚓长度减去q，期望得分85pts

进一步观察，我们发现，同一种切法，早切的蚯蚓一定比晚切的长，根据这个性质，维护三个单调队列，分别维护未被切割的蚯蚓长度，被切割过的蚯蚓长度中较长的那些的长度，被切割的蚯蚓长度中较短的那些的长度，时间复杂度O(M),可以通过所有数据

[代码]

注意为了避免精度误差，需使用long double类型，否则可能无法通过UOJ Extra Test

#include<bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const long long MAXN = 1e7;

long long i,n,m,u,value,t,x,y,mx,pos,v;
long long head[],tail[],ans1[MAXN],ans2[MAXN];
long long q[][MAXN];

template <typename T> inline void read(T &x)
{
        long long f = ; x = ;
        char c = getchar();
        for (; !isdigit(c); c = getchar())
        {
                if (c == '-') f = -f;
        }
        for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
        x *= f;
}

int main()
{

        read(n); read(m); read(value); read(u); read(v); read(t);
        for (i = ; i <= n; i++)
        {
                read(x);
                q[][++tail[]] = x;
        }
        head[] = head[] = head[] = ;
        sort(q[] + ,q[] + tail[] + ,greater<long long>());
        for (i = ; i <= m; i++)
        {
                mx = -INF;
                if (head[] <= tail[] && q[][head[]] > mx)
                {
                        mx = q[][head[]];
                        pos = ;
                }
                if (head[] <= tail[] && q[][head[]] > mx)
                {
                        mx = q[][head[]];
                        pos = ;
                }
                if (head[] <= tail[] && q[][head[]] > mx)
                {
                        mx = q[][head[]];
                        pos = ;
                }
                ans1[i] = mx + (i - ) * value;
                head[pos]++;
                x = (long long)((mx + (i - ) * value) * (long double)1.0 * u / v) - i * value;
                y = mx + (i - ) * value - (long long)((mx + (i - ) * value) * (long double)1.0 * u / v) - i * value;
                q[][++tail[]] = max(x,y);
                q[][++tail[]] = min(x,y);
        }
        for (i = ; i * t <= m; i++) printf("%lld ",ans1[i * t]);
        printf("\n");
        for (i = ; i <= n + m; i++)
        {
                mx = -INF;
                if (head[] <= tail[] && q[][head[]] > mx)
                {
                        mx = q[][head[]];
                        pos = ;
                }
                if (head[] <= tail[] && q[][head[]] > mx)
                {
                        mx = q[][head[]];
                        pos = ;
                }
                if (head[] <= tail[] && q[][head[]] > mx)
                {
                        mx = q[][head[]];
                        pos = ;
                }
                ans2[i] = mx + m * value;
                head[pos]++;
        }
        for (i = ; i * t <= n + m; i++) printf("%lld ",ans2[i * t]);
        printf("\n");

        return ;

}