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  • [BZOJ4993||4990] [Usaco2017 Feb]Why Did the Cow Cross the Road II(DP + 线段树)

    传送门

    f[i][j]表示当前第i个,且最后一个位置连接到j

    第一维可以省去,能连边的点可以预处理出来,dp可以用线段树优化

    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    #define N 100001
    #define root 1, 1, n
    #define ls now << 1, l, mid
    #define rs now << 1 | 1, mid + 1, r
    
    using namespace std;
    
    int n, ans, cnt;
    int a[N], pos[N], sum[N << 2], f[N];
    
    inline int read()
    {
    	int x = 0, f = 1;
    	char ch = getchar();
    	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    	return x * f;
    }
    
    inline int query(int now, int l, int r, int x, int y)
    {
    	if(x > y) return 0;
    	if(x <= l && r <= y) return sum[now];
    	int mid = (l + r) >> 1, ret = 0;
    	if(x <= mid) ret = max(ret, query(ls, x, y));
    	if(mid < y) ret = max(ret, query(rs, x, y));
    	return ret;
    }
    
    inline void update(int now, int l, int r, int x, int d)
    {
    	if(l == r)
    	{
    		sum[now] = max(sum[now], d);
    		return;
    	}
    	int mid = (l + r) >> 1;
    	if(x <= mid) update(ls, x, d);
    	else update(rs, x, d);
    	sum[now] = max(sum[now << 1], sum[now << 1 | 1]);
    }
    
    int main()
    {
    	int i, j, x;
    	n = read();
    	for(i = 1; i <= n; i++) x = read(), pos[x] = i;
    	for(i = 1; i <= n; i++)
    	{
    		cnt = 0;
    		x = read();
    		for(j = max(1, x - 4); j <= min(n, x + 4); j++) a[++cnt] = pos[j];
    		sort(a + 1, a + cnt + 1);
    		for(j = cnt; j >= 1; j--)
    		{
    			x = query(root, 1, a[j] - 1) + 1;
    			f[a[j]] = max(f[a[j]], x);
    			update(root, a[j], x);
    		}
    	}
    	for(i = 1; i <= n; i++) ans = max(ans, f[i]);
    	printf("%d
    ", ans);
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhenghaotian/p/7568650.html
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