[SDOI2012]任务安排

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=2726

[算法]

        此题与POJ1180非常相似 

        但是 ， 此题中的t值可能为负 ， 这意味着不能每次都将斜率 <= k的点弹出 ， 而需要在凸壳中进行二分查找

        时间复杂度 : O(NlogN)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

int n , S , l , r;
ll sumc[N] , sumt[N] , f[N];
int q[N] , c[N] , t[N];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T 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;
}
inline int _binary_search(int L , int R , int val)
{
        int l = L , r = R , ret = L;
        while (l <= r)
        {
                int mid = (l + r) >> 1;
                if (f[q[mid + 1]] - f[q[mid]] <= val * (sumc[q[mid + 1]] - sumc[q[mid]]))
                {
                        ret = mid + 1;
                        l = mid + 1;        
                }    else r = mid - 1;
        }        
        return ret;
}

int main()
{
        
        read(n); read(S);
        for (int i = 1; i <= n; i++)
        {
                read(t[i]);
                read(c[i]);
                sumt[i] = sumt[i - 1] + t[i];
                sumc[i] = sumc[i - 1] + c[i];
        }
        f[q[l = r = 1] = 0] = 0;
        for (int i = 1; i <= n; i++)
        {
                int pos = _binary_search(l , r - 1 , S + sumt[i]);
                f[i] = f[q[pos]] - sumc[q[pos]] * (S + sumt[i]) + sumt[i] * sumc[i] + S * sumc[n];
                while (l < r && (f[i] - f[q[r]]) * (sumc[q[r]] - sumc[q[r - 1]]) <= (f[q[r]] - f[q[r - 1]]) * (sumc[i] - sumc[q[r]])) --r;
                q[++r] = i;         
        }
        printf("%lld
" , f[n]);
        
        return 0;
    
}