TJU ACM-ICPC Online Judge—1191 The Worm Turns

B - The Worm Turns

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

Description

Worm is an old computer game. There are many versions, but all involve maneuvering a "worm" around the screen, trying to avoid running the worm into itself or an obstacle.

We'll simulate a very simplified version here. The game will be played on a 50 x 50 board, numbered so that the square at the upper left is numbered (1, 1). The worm is initially a string of 20 connected squares. Connected squares are adjacent horizontally or vertically. The worm starts stretched out horizontally in positions (25, 11) through (25, 30), with the head of the worm at (25, 30). The worm can move either East (E), West (W), North (N) or South (S), but will never move back on itself. So, in the initial position, a W move is not possible. Thus the only two squares occupied by the worm that change in any move are its head and tail. Note that the head of the worm can move to the square just vacated by the worm's tail.

You will be given a series of moves and will simulate the moves until either the worm runs into itself, the worm runs off the board, or the worm successfully negotiates its list of moves. In the first two cases you should ignore the remaining moves in the list.

Input

There will be multiple problems instances. The input for each problem instance will be on two lines. The first line is an integer n (<100) indicating the number of moves to follow. (A value of n = 0 indicates end of input.) The next line contains n characters (either E, W, N or S), with no spaces separating the letters, indicating the sequence of moves.

Output

Generate one line of output for each problem instance. The output line should be one of the follow three:

The worm ran into itself on move m.

The worm ran off the board on move m.

The worm successfully made all m moves.

Where m is for you to determine and the first move is move 1.

Sample Input

18

NWWWWWWWWWWSESSSWS

20

SSSWWNENNNNNWWWWSSSS

30

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

13

SWWWWWWWWWNEE

0

Sample Output

The worm successfully made all 18 moves.

The worm ran into itself on move 9.

The worm ran off the board on move 21.

The worm successfully made all 13 moves.

这个题目的话，相信大部分同学都能一眼看出其意思，即使是一个英文题，哈哈，这不就是我们玩的贪吃蛇游戏，判断死亡的一个方法吗，题目不是很难，话不多说看代码先：

AC代码如下：（如有错误和建议，请大家不吝指出）

#include<stdio.h>
#include<string.h>
int main() {
	int n,hgs[55][55],wx,wy,tx,ty;
	int flag1,flag2;
	char str[105];
	while(scanf("%d",&n),n) {
		getchar();
		scanf("%s",str);
		memset(hgs,0,sizeof(hgs));//每次游戏开始重置归零
		for(int i=11; i<=30; i++)	hgs[25][i]=1;
		tx=25;	ty=30;
		wx=25;	wy=11;
		flag1=flag2=0;
		int k=0;
		int count=0;
		for(int i=0;; i++) {
			if(str[i]==' ')	continue;  //还是判断一下有没有空格的好，被坑怕了，小心一点为好
			//先处理尾部的运动
			if(count+1<20) {
				hgs[wx][wy]=0;	wy++;
			} else {
				hgs[wx][wy]=0;
				while(str[k]==' ') 	 k++;
				if(str[k]=='E')   wy++;
				else if(str[k]=='S') 	wx--;
				else if(str[k]=='W') 	wy--;
				else if(str[k]=='N') 	wx++;
				k++;
			}

			//后处理头部的运动
			if(str[i]=='E') {  //向右走
				ty++;
				if(ty>50) {
					flag1=1;  break;
				}
				if(hgs[tx][ty]==0)	hgs[tx][ty]=1;
				else {
					flag2=1;  break;   //头撞到了自己
				}
			} else if(str[i]=='S') {   //向下走
				tx--;
				if(tx<1) {
					flag1=1;  break;
				}
				if(hgs[tx][ty]==0)  hgs[tx][ty]=1;
				else {
					flag2=1;  break;
				}
			} else if(str[i]=='W') {  //向左走
				ty--;
				if(ty<1) {
					flag1=1;  break;
				}
				if(hgs[tx][ty]==0)	hgs[tx][ty]=1;
				else {
					flag2=1;  break;
				}
			} else if(str[i]=='N') {   //向上走
				tx++;
				if(tx>50) {
					flag1=1;  break;
				}
				if(hgs[tx][ty]==0)	hgs[tx][ty]=1;
				else {
					flag2=1;  break;
				}
			}
			count++;
			if(count==n)  break;
		}
		if(count==n)  printf("The worm successfully made all %d moves.\n",count);
		else if(flag1)	printf("The worm ran off the board on move %d.\n",count+1);
		else if(flag2)  printf("The worm ran into itself on move %d.\n",count+1);
	}
	return 0;
}