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≤iN−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:

N
s

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

6
ooxoox

Sample Output 1

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

3
oox

Sample Output 2

-1

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


Sample Input 3

10
oxooxoxoox

Sample Output 3

SSWWSSSWWS

题目链接:http://arc069.contest.atcoder.jp/tasks/arc069_b

题意:n只动物从1到n围成一个圈,每只动物要么是羊要么是狼。每只动物会说出一个字母,说'o'表示它两边动物种类相同,说'x'表示不同。但羊是说真话,狼是说反话。求出这n只动物的种类。
分析:模拟一下就可以了,不过这个模拟比较大!
下面给出AC代码:
 #include <bits/stdc++.h>
using namespace std;
const int N=;
int gh(char *str,char *ch,int n) {
for(int i=; i<n; i++) {
if(i<n-) {
if(str[i]=='o') {
if(ch[i]=='S') {
ch[i+]=ch[i-];
} else {
if(ch[i-]=='S') ch[i+]='W';
else ch[i+]='S';
}
} else {
if(ch[i]=='S') {
if(ch[i-]=='S') ch[i+]='W';
else ch[i+]='S';
} else {
ch[i+]=ch[i-];
}
}
}
else if(i==n-) {
if(str[i]=='o') {
if(ch[i]=='S') {
if(ch[i+]!=ch[i-]) {
return ;
}
}
else {
if(ch[i+]==ch[i-]){
return ;
}
}
}
else{
if(ch[i]=='S'){
if(ch[i+]==ch[i-]){
return ;
}
}
else{
if(ch[i+]!=ch[i-]) {
return ;
}
}
}
}
else if(i==n-){
if(str[i]=='x'){
if(ch[i]=='S') {
if(ch[n-]==ch[]) return ;
}
else{
if(ch[n-]!=ch[]) return ;
}
}
else{
if(ch[i]=='S'){
if(ch[n-]!=ch[]) return ;
}
else{
if(ch[n-]==ch[]) return ;
}
}
}
}
return ;
}
int main() {
char str[N];
char ch[N];
int n;
int flag=;
scanf("%d",&n);
scanf("%s",str);
ch[]='S';
if(str[]=='o') {
memset(ch,,sizeof(ch));
ch[]='S';
ch[]='S';
ch[n-]='S';
flag=gh(str,ch,n);
if(flag==) {
for(int i=; i<n; i++) printf("%c",ch[i]);
puts("");
return ;
} else {
memset(ch,,sizeof(ch));
ch[]='S';
ch[]='W';
ch[n-]='W';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
return ;
}
}
}
else{
memset(ch,,sizeof(ch));
ch[]='S';
ch[]='S';
ch[n-]='W';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
return ;
}
else{
memset(ch,,sizeof(ch));
ch[]='S';
ch[]='W';
ch[n-]='S';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
return ;
}
}
}
ch[]='W';
if(str[]=='o'){
memset(ch,,sizeof(ch));
ch[]='W';
ch[]='S';
ch[n-]='W';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
return ;
}
else{
ch[]='W';
ch[]='W';
ch[n-]='S';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
return ;
}
}
}
else{
memset(ch,,sizeof(ch));
ch[]='W';
ch[]='S';
ch[n-]='S';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
}
else{
ch[]='W';
ch[]='W';
ch[n-]='W';
flag=gh(str,ch,n);
if(flag==){
for(int i=;i<n;i++) printf("%c",ch[i]);
puts("");
return ;
}
}
}
puts("-1");
return ;
}

AtCoder Regular Contest 069 D的更多相关文章

  1. AtCoder Regular Contest 069 F Flags 二分,2-sat,线段树优化建图

    AtCoder Regular Contest 069 F Flags 二分,2-sat,线段树优化建图 链接 AtCoder 大意 在数轴上放上n个点,点i可能的位置有\(x_i\)或者\(y_i\ ...

  2. AtCoder Regular Contest 069 D - Menagerie 枚举起点 模拟递推

    arc069.contest.atcoder.jp/tasks/arc069_b 题意:一堆不明身份的动物排成一圈,身份可能是羊或狼,羊一定说实话,狼一定说假话.大家各自报自己的两边是同类还是不同类, ...

  3. AtCoder Regular Contest 069 F - Flags

    题意: 有n个点需要摆在一个数轴上,每个点需要摆在ai这个位置或者bi上,问怎么摆能使数轴上相邻两个点之间的距离的最小值最大. 二分答案后显然是个2-sat判定问题,因为边很多而连边的又是一个区间,所 ...

  4. AtCoder Regular Contest 069

    1. C - Scc Puzzle 计算scc的个数,先判断s个数需要多少个cc,多的cc,每四个可以组成一个scc.注意数据范围,使用long long. #include<bits/stdc ...

  5. AtCoder Regular Contest 061

    AtCoder Regular Contest 061 C.Many Formulas 题意 给长度不超过\(10\)且由\(0\)到\(9\)数字组成的串S. 可以在两数字间放\(+\)号. 求所有 ...

  6. AtCoder Regular Contest 094 (ARC094) CDE题解

    原文链接http://www.cnblogs.com/zhouzhendong/p/8735114.html $AtCoder\ Regular\ Contest\ 094(ARC094)\ CDE$ ...

  7. AtCoder Regular Contest 092

    AtCoder Regular Contest 092 C - 2D Plane 2N Points 题意: 二维平面上给了\(2N\)个点,其中\(N\)个是\(A\)类点,\(N\)个是\(B\) ...

  8. AtCoder Regular Contest 093

    AtCoder Regular Contest 093 C - Traveling Plan 题意: 给定n个点,求出删去i号点时,按顺序从起点到一号点走到n号点最后回到起点所走的路程是多少. \(n ...

  9. AtCoder Regular Contest 094

    AtCoder Regular Contest 094 C - Same Integers 题意: 给定\(a,b,c\)三个数,可以进行两个操作:1.把一个数+2:2.把任意两个数+1.求最少需要几 ...

随机推荐

  1. 1. 生成三行文本,过滤文本,cp不覆盖,find查找文件,sed打印行,查看系统信息,磁盘分区

  2. LINUX:alias命令详解

    发现目前安装的g++并没有开启选项 -std=c++11,无法使用c++11的新标准及其中的列表初始化.搜索后得到解决方法:键入:alias  g++="g++ -std=c++11&quo ...

  3. ArcGIS API for JavaScript 4.2学习笔记[14] 弹窗的位置、为弹窗添加元素

    这一节我们来看看弹窗的位置和弹窗上能放什么. 先一句话总结: 位置:可以随便(点击时出现或者一直固定在某个位置),也可以指定位置 能放什么:四种,文字.媒体(图片等).表格.附件. [Part I 位 ...

  4. 入门级Nginx反向代理nodejs

    本着想实现前后端分离开发的初衷,我决定学习一下关于nignx反向代理的配置. 1.下载Nginx稳定版本 2.打开nginx配置文件 nginx.conf: 3.在http模块的server部分配置 ...

  5. Docker(八):Docker端口映射

    1.随机映射 docker run -P -d --name mynginx1 nginx [root@node1 ~]# docker ps -l CONTAINER ID IMAGE COMMAN ...

  6. java 操作本地数据库 mysql

    单线程版 /** * */ import java.sql.*; import java.util.Date; import org.omg.CORBA.PUBLIC_MEMBER; /** * @a ...

  7. 关于理解python类的小题

    今天看了python部落翻译的一篇<一道python类的小题>文章,感觉挺有启发性,记录下来: print('A') class Person(object): print('B') de ...

  8. HTML中的超链接

    超链接:也叫URL(Uniform Resource Locator),就是统一资源定位器.一般效果是我们点击网页上某个地方,网页会自动跳转到另外一个地方. 一般链接遵循以下要求:scheme://h ...

  9. springBoot系列教程08:拦截器(Interceptor)的使用

    拦截器intercprot  和 过滤器 Filter 其实作用类似 在最开始接触java 使用struts2的时候,里面都是filter 后来springmvc时就用interceptor 没太在意 ...

  10. Q:记学习枚举过程中的一个小问题

    在学习有关java枚举的时候,我们知道了所有的枚举类型均是继承自java.lang.Enum类的,且所有的枚举常量均是该枚举类型的一个对象,且对象名即为该枚举常量的名称.例子如下:源码: public ...