Atcoder 2334 D

题目链接：http://abc055.contest.atcoder.jp/tasks/arc069_b?lang=en

D - Menagerie

---

Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

Snuke, who loves animals, built a zoo.

There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle. The animal numbered i(2≤i≤N−1) is adjacent to the animals numbered i−1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N−1 and 1.

There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies.

Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered si. Here, if si is o, the animal said that the two neighboring animals are of the same species, and if si is x, the animal said that the two neighboring animals are of different species.

More formally, a sheep answered o if the two neighboring animals are both sheep or both wolves, and answered x otherwise. Similarly, a wolf answered x if the two neighboring animals are both sheep or both wolves, and answered o otherwise.

Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print -1.

Constraints

• 3≤N≤105
• s is a string of length N consisting of o and x.

---

Input

The input is given from Standard Input in the following format:

Ns

Output

If there does not exist an valid assignment that is consistent with s, print -1. Otherwise, print an string t in the following format. The output is considered correct if the assignment described by t is consistent with s.

• t is a string of length N consisting of S and W.
• If ti is S, it indicates that the animal numbered i is a sheep. If ti is W, it indicates that the animal numbered i is a wolf.

---

Sample Input 1

Copy

6ooxoox

Sample Output 1

Copy

SSSWWS

For example, if the animals numbered 1, 2, 3, 4, 5 and 6 are respectively a sheep, sheep, sheep, wolf, wolf, and sheep, it is consistent with their responses. Besides, there is another valid assignment of species: a wolf, sheep, wolf, sheep, wolf and wolf.

Let us remind you: if the neiboring animals are of the same species, a sheep answers o and a wolf answers x. If the neiboring animals are of different species, a sheep answers x and a wolf answers o.

---

Sample Input 2

Copy

3oox

Sample Output 2

Copy

-1

Print -1 if there is no valid assignment of species.

---

Sample Input 3

Copy

10oxooxoxoox

Sample Output 3

Copy

SSWWSSSWWS

---

Submit

题意：动物园中只有狼和羊， 狼羊做成一个圈，输入一个字符串，如果是羊，表示真话，串中'o'表示是它的左右为同一物种，‘x’表示不同物种，如果是狼则相反，，问是否存在一种可能，满足上面的字符串，最后输出1-n位置的狼和羊

解析：刚看到这个题时，完全懵逼，这TM怎么模拟，后来看泰神博客才知道怎么写，就是枚举前两个动物，得出一个序列，判断下是否成立，成立的话输出即可，复杂度为4*n，就是O(n)

泰神博客：http://www.wonter.net/

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<map>#include<cmath>#define N 100009using namespace std;const int INF = 0x3f3f3f3f;char s[N];int ans[N];int nx[4] = {1, 1, 0, 0};int ny[4] = {1, 0, 0, 1};bool check(int a, int b, int n){    ans[0] = a; ans[1] = b;    for(int i = 1; i < n - 1; i++)    {        if(s[i] == 'o')        {            if(ans[i]) ans[i + 1] = ans[i - 1];            else ans[i + 1] = !ans[i - 1];        }        else        {            if(ans[i]) ans[i + 1] = !ans[i - 1];            else ans[i + 1] = ans[i - 1];        }    }    if(s[0] == 'o')    {        if(ans[0])        {            if(ans[1] != ans[n - 1]) return false;        }        else if(ans[1] == ans[n - 1]) return false;    }    else    {        if(ans[0])        {            if(ans[1] == ans[n - 1]) return false;        }        else if(ans[1] != ans[n - 1]) return false;    }    if(s[n - 1] == 'o')    {        if(ans[n - 1])        {            if(ans[0] != ans[n - 2]) return false;        }        else if(ans[0] == ans[n - 2]) return false;    }    else    {        if(ans[n - 1])        {            if(ans[0] == ans[n - 2]) return false;        }        else if(ans[0] != ans[n - 2]) return false;    }    return true;}int main(){    int n;    scanf("%d %s", &n, s);    for(int i = 0; i < 4; i++)    {        if(check(nx[i], ny[i], n))        {            for(int i = 0; i < n; i++)            {                if(ans[i]) printf("S");                else printf("W");            }            printf("\n");            exit(0);        }    }    puts("-1");    return 0;}