P2205 [USACO13JAN]画栅栏Painting the Fence 题目描述 Farmer John has devised a brilliant method to paint the long fence next to his barn (think of the fence as a one-dimensional number line). He simply attaches a paint brush to his favorite cow Bessie, and then retires to drink a cold ...
#include<cstdio>#include<cstdlib>#include<algorithm>#definemaxn 200010usingnamespacestd;inta[maxn],b[maxn],cnt=0,idx=0;structadd{intpos,val;add(){}add(intx,inty){pos=x,val=y;}booloperator<(add t){returnpos<t.pos; } } delta[maxn];intmain(){inti,j,n,m,k,x,p=0,ans=0;ch...
Fence Paintingwww.usaco.org/index.php?page=viewproblem2&cpid=567 import sys sys.stdin = open('paint.in', "r") sys.stdout = open('paint.out', "w") a, b = map(int, input().split()) c, d = map(int, input().split()) s = set() for i in range(a, b): s.add(i...
将区间的每个端点都保留下来,从小到大排序后,维护涂了涂料的个数 ,如果 的话就记录下这段区间的数。 类似:传送门 代码: // Problem: P2205 [USACO13JAN]Painting the Fence S // Contest: Luogu // URL: https://www.luogu.com.cn/problem/P2205 // Memory Limit: 125 MB // T...
离散-洛谷P2205 [USACO13JAN]画栅栏Painting the Fence https://daniu.luogu.org/problem/show?pid=2205 本来要抄题解的 ,结果发现搜不到; …. 其实就是一个离散化; 假如现在的点是now; 要向右移动x个单位; 我们直接在now的地方+1; 在now+x的地方-1; 就好了; 然后排序判重乱搞; 要注意它是一个...
Please determine the largest size of a fair photo FJ can take, given that FJ has the option of painting some of his white cows (of course, he does not need to paint any of the white cows if he decides this is better). 可以先任意把0染成1. 区间长度定义为,[L,R]中最右和最左的数...
倒着扫一遍,因为某头牛后面的牛对这头牛的速度没影响。 计算出ts后牛的终点,如果能撞上前面最慢的牛,那么小团体数+1 注意开long long 一开始不理解为什么倒着扫, 因为如果正着扫,看第i头牛能否撞上i+1头, 我们不确定第i+1头牛的速度,可能第i+1头牛 ...
接下来n行描述每次操作的步数和方向 Sample Input 6 2 2 R 6 L 1 R 8 L 1 R 2 R Sample Output 6 下面的方法用了 map 和 区间表示法 http://www.cnblogs.com/zquzjx/p/8321466.html(区间表示法看这里) 但是不同于 题解https://www.luogu.org/problemnew/solution/P2205扫描线的方法可以得到整段...
洛谷P2205 [USACO13JAN]画栅栏Painting the Fence 这是我的做法,我用了离散化+二分,比较慢 1#include <bits/stdc++.h>2#defineFor(i,j,k) for(int i=j;i<=k;i++)3#defineint long long4#defineLL long long5usingnamespacestd ;67constintN =100011;8structnode{9intl,r,del ;10}cnt[N*2]...
没有询问的区间修改,差分做就好了。 代码: #include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#defineN 10009usingnamespacestd;intn,I,H,R;intC[N];structQ{intl,r; }f[N];boolcmp(Q a,Q b){if(a.l==b.l)returna.r<b.r;returna.l<b.l; ...