USACO Section 1.1.7 Broken Necklace

题目 

(由于USACO在外部无法直接访问题库,所以把题目贴上,可以通过上面的目录跳过)

<center style="font&#45;size&#58;14px&#59;">

Broken Necklace

</center>

You have a necklace of N red, white, or blue beads (3<=N<=350) some of which are red, others blue, and others white, arranged at random. Here are two examples for n=29:

                1 2                               1 2
            r b b r                           b r r b
          r         b                       b         b
         r           r                     b           r
        r             r                   w             r
       b               r                 w               w
      b                 b               r                 r
      b                 b               b                 b
      b                 b               r                 b
       r               r                 b               r
        b             r                   r             r
         b           r                     r           r
           r       r                         r       b
             r b r                             r r w
            Figure A                         Figure B
                        r red bead
                        b blue bead
                        w white bead

The beads considered first and second in the text that follows have been marked in the picture.

The configuration in Figure A may be represented as a string of b's and r's, where b represents a blue bead and r represents a red one, as follows: brbrrrbbbrrrrrbrrbbrbbbbrrrrb .

Suppose you are to break the necklace at some point, lay it out straight, and then collect beads of the same color from one end until you reach a bead of a different color, and do the same for the other end (which might not be of the same color as the beads collected before this).

Determine the point where the necklace should be broken so that the most number of beads can be collected.

Example

For example, for the necklace in Figure A, 8 beads can be collected, with the breaking point either between bead 9 and bead 10 or else between bead 24 and bead 25.

In some necklaces, white beads had been included as shown in Figure B above. When collecting beads, a white bead that is encountered may be treated as either red or blue and then painted with the desired color. The string that represents this configuration will include the three symbols r, b and w.

Write a program to determine the largest number of beads that can be collected from a supplied necklace.

PROGRAM NAME: beads

INPUT FORMAT

Line 1: N, the number of beads
Line 2: a string of N characters, each of which is r, b, or w

SAMPLE INPUT (file beads.in)

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

OUTPUT FORMAT

A single line containing the maximum of number of beads that can be collected from the supplied necklace.

SAMPLE OUTPUT (file beads.out)

11

OUTPUT EXPLANATION

Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.
                       original 'split'
                             v
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb|wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
                       ******|*****
                       rrrrrb|bbbbb  <-- assignments
                       5 x r  6 x b  <-- 11 total


思路

枚举

先看数据规模,n=350,O(n2)无压力。所以可以枚举每个断点,然后算出能拾几个珠子。貌似不是O(n2),只是最坏的情况而已。应该也没问题吧,没试SORRY~

换个神似靠谱的法子吧

其实最开始,把题目错搞成找一串颜色相同的最长子串。想的是扫描一遍,对于每个点只需要判断前后能不能接上然后找出最长的就行了。后来发现其实要从一个断点向两头去寻找,但是全局代码懒得改了,就这样继续了。

还是从头到尾扫描找出每个断点向左能挂几个,从后往前扫描统计每个断点能向右挂几个。对于每一个珠子有三种情况:

1.跟前面一样或者自己是白色的,直接挂上;

2.剩下的就是另一种颜色了(既跟前面不一样又还不是白的)。这样的话首先要判断前面是不是白色的

1)如果是白色的还得把前面的颜色一样的也挂。

2)如果还不是白色的话只能另起炉灶。当然此时要先看一下挂了多长要不要替换answer。

PK

改完之后发现代码好长好长,而且核心代码重复性很大,第一种方法其实会更好的,但是要注意第一种方法还是要从每个点去向两头搜,往两头搜还是会有很多重复代码。感觉这两种法子没有什么大区别?

代码

#include <cstdio>
char a[777];
int l[777],r[777],n;
int main(){
	freopen("beads.in","r",stdin);
	freopen("beads.out","w",stdout);
	scanf("%d%s",&n,a);
	for (int i(0);i<n;i++){
		a[n+i]=a[i];
	}
	int i=0,j=a[0];
	while (++i<2*n){
		if ( a[i]==j || a[i]=='w' ){
			l[i]=l[i-1]+1;
		}
		else {
			if (a[i-1]=='w'){
				int k=i-1;
				while (k>=0 && a[k]=='w'){k--;}
				l[i]=i-k-1;
			}
			j=a[i];
		}
	}
	i=2*n-1;
	j=a[2*n-1];
	while (--i>=0){
		if ( a[i]==j || a[i]=='w'){
			r[i]=r[i+1]+1;
		}
		else {
			if (a[i+1]=='w'){
				int k=i+1;
				while (k<2*n && a[k]=='w'){k++;}
				r[i]=k-i-1;
			}
			j=a[i];
		}
	}
	for (int i(0);i<n;i++) l[i]=l[i+n];
	int ans=0;
	for (int i(0);i<n-1;i++){
		if (l[i]+r[i+1]>ans) ans=l[i]+r[i+1];
	}
	if (ans>n){
		printf("%d\n",n);
	} else {
		printf("%d\n",ans+2);
	}
	return 0;
}



注意!此信息未认证,请谨慎判断信息的真实性!

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