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  • Solution -「AGC 034C」Tests

    (mathcal{Description})

      Link.

      给定非负整数序列 ({l_n},{r_n},{b_n},X),求最小的 (s),使得存在非负整数序列 ({a_n},{c_n}),满足 (a_ile X)(sum_{i=1}^na_i=s)(c_iin[l_i,r_i]),且

    [sum_{i=1}^nc_i(a_i-b_i)ge0 ]

      所有输入均 (le10^5)

    (mathcal{Solution})

      显然二分 (s),仅需做到检测某个 (s) 是否合法。下令 (w=sum_{i=1}^nc_ia_i)

      假设 ({a_n}) 已经确定,那么所有满足 (a_ige b_i)(c_i=r_i),其余 (c_i=l_i)。考虑初始时所有 (a_i=0,c_i=l_i),现在把 (s)(1) 挨个加到一些 (a_i) 上。当 (a_i<b_i) 时,对 (w) 贡献 (l_i)(此时 (c_i) 仍取 (l_i));当 (a_i=b_i) 时,对 (w) 贡献由 (l_i) 转为 (r_i)(c_i) 变成 (r_i));继续增加,对 (w) 贡献 (r_i)。最终仅需比较 (w)(sum_{i=1}^nl_ib_i) 的大小。

      所以问题抽象为:有 (n) 个分段函数 (f_{1..n}(x)),满足

    [f_i(x)=egin{cases}l_ix&xin[0,b_i]capmathbb N\ l_ib_i+r_i(x-b_i)&xin(b_i,X]capmathbb N end{cases} ]

      仅需钦定 ({x_n}),使得 (sum_{i=1}^nf_i(x_i)) 取最大。

      考虑贪心,不难证明:至多有一个 (0<x_i<X)。直接枚举哪一个 (0<x_i<X),贪心地选取最大的 (f_i(X)),即可 (mathcal O(n)) 检测。最终复杂度 (mathcal O(nlogsum_{i=1}^nb_i))

    (mathcal{Code})

    /* Clearink */
    
    #include <cstdio>
    #include <algorithm>
    
    #define rep( i, l, r ) for ( int i = l, repEnd##i = r; i <= repEnd##i; ++i )
    #define per( i, r, l ) for ( int i = r, repEnd##i = l; i >= repEnd##i; --i )
    
    inline int rint () {
    	int x = 0, f = 1; char s = getchar ();
    	for ( ; s < '0' || '9' < s; s = getchar () ) f = s == '-' ? -f : f;
    	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
    	return x * f;
    }
    
    typedef long long LL;
    
    const int MAXN = 1e5;
    int n, x, b[MAXN + 5], l[MAXN + 5], r[MAXN + 5], ord[MAXN + 5];
    LL full[MAXN + 5];
    
    inline LL contr ( const int i, const int s ) {
    	return s <= b[i] ? 1ll * s * l[i]
    		: 1ll * l[i] * b[i] + 1ll * ( s - b[i] ) * r[i];
    }
    
    inline void init () {
    	std::sort ( ord + 1, ord + n + 1, []( const int i, const int j ) {
    		return contr ( i, x ) > contr ( j, x );
    	} );
    	rep ( i, 1, n ) full[i] = full[i - 1] + contr ( ord[i], x );
    }
    
    inline LL calc ( const LL scr ) {
    	/*
    	 * let's come up with a greedy algorithm!
    	 * */
    	int fcnt = scr / x, rest = scr % x;
    	LL ret = 0;
    	rep ( i, 1, n ) { // score on exam <ord[i]> is <rest>.
    		LL cur = contr ( ord[i], rest );
    		if ( i > fcnt ) cur += full[fcnt];
    		else cur += full[fcnt + 1] - contr ( ord[i], x );
    		ret = cur > ret ? cur : ret;
    	}
    	return ret;
    }
    
    int main () {
    	n = rint (), x = rint ();
    	LL sum = 0, sb = 0;
    	rep ( i, 1, n ) {
    		ord[i] = i;
    		b[i] = rint (), l[i] = rint (), r[i] = rint ();
    		sum += 1ll * b[i] * l[i], sb += b[i];
    	}
    	init ();
    	LL lef = 0, rig = sb;
    	while ( lef < rig ) {
    		LL mid = lef + rig >> 1;
    		if ( calc ( mid ) >= sum ) rig = mid;
    		else lef = mid + 1;
    	}
    	printf ( "%lld
    ", lef );
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/rainybunny/p/14284256.html
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