[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[10],tail[10],ans1[MAXN],ans2[MAXN];
long long q[3][MAXN];

template <typename T> inline void read(T &x)
{
        long long f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar())
        {
                if (c == '-') f = -f;
        }
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}

int main() 
{
        
        read(n); read(m); read(value); read(u); read(v); read(t);
        for (i = 1; i <= n; i++)
        {
                read(x);
                q[0][++tail[0]] = x; 
        } 
        head[0] = head[1] = head[2] = 1;
        sort(q[0] + 1,q[0] + tail[0] + 1,greater<long long>());
        for (i = 1; i <= m; i++)
        {
                mx = -INF;
                if (head[0] <= tail[0] && q[0][head[0]] > mx)
                {
                        mx = q[0][head[0]];
                        pos = 0;
                }
                if (head[1] <= tail[1] && q[1][head[1]] > mx)
                {
                        mx = q[1][head[1]];
                        pos = 1;
                }
                if (head[2] <= tail[2] && q[2][head[2]] > mx)
                {
                        mx = q[2][head[2]];
                        pos = 2;
                }
                ans1[i] = mx + (i - 1) * value;
                head[pos]++;
                x = (long long)((mx + (i - 1) * value) * (long double)1.0 * u / v) - i * value;
                y = mx + (i - 1) * value - (long long)((mx + (i - 1) * value) * (long double)1.0 * u / v) - i * value;
                q[1][++tail[1]] = max(x,y);
                q[2][++tail[2]] = min(x,y);
        }
        for (i = 1; i * t <= m; i++) printf("%lld ",ans1[i * t]);
        printf("\n");
        for (i = 1; i <= n + m; i++) 
        {
                mx = -INF;
                if (head[0] <= tail[0] && q[0][head[0]] > mx)
                {
                        mx = q[0][head[0]];
                        pos = 0;
                }
                if (head[1] <= tail[1] && q[1][head[1]] > mx)
                {
                        mx = q[1][head[1]];
                        pos = 1;
                }
                if (head[2] <= tail[2] && q[2][head[2]] > mx)
                {
                        mx = q[2][head[2]];
                        pos = 2;
                }
                ans2[i] = mx + m * value;
                head[pos]++;
        }
        for (i = 1; i * t <= n + m; i++) printf("%lld ",ans2[i * t]);
        printf("\n");
        
        return 0;
    
}