URAL 2027 URCAPL, Episode 1 (模拟)

题意:给你一个HxW的矩阵,每个点是一个指令,根据指令进行一系列操作。

题解:模拟

 

#include<cstdio>
#include<algorithm>
using namespace std;


const int maxn = 101;
char G[maxn][maxn];

int dx[] = {-1,0,1, 0};
int dy[] = { 0,1,0,-1};

struct pointer
{
    int r,c;
    int dir;
    void Move(){
        r += dx[dir];
        c += dy[dir];
    }
    char read(){
        return G[r][c];
    }
}p;

    int h,w;
const int up_bound = 1e5;
const int TimeLim = 1e6;
const int MaxN = 1e5+5;
int readLim;
int readList[MaxN];

void execute()
{
    p.r = p.c = 0;
    int Register[26] = {0};
    int cur = 0;
    char op = p.read();
    p.dir = 1;
    int step = 1;
    int ReadTimes = 0;
   while(op!='#'){
        if('A'<=op&&op<='Z'){
            swap(Register[op-'A'],cur);
        }else switch(op){
                case '^':{
                    p.dir = 0;
                    break;
                }
                case '>':{
                    p.dir = 1;
                    break;
                }
                case 'v':{
                    p.dir = 2;
                    break;
                }
                case '<':{
                    p.dir = 3;
                    break;
                }
                case '?':{
                    if(ReadTimes>=readLim){
                        cur = readList[readLim-1];
                    }else {
                        cur = readList[ReadTimes];
                        ReadTimes++;
                    }
                    break;
                }
                case '!':{
                    printf("%d\n",cur);
                    cur = 0;
                    break;
                }
                case '+':{
                    cur++;
                    if(abs(cur)>up_bound){
                        printf("OVERFLOW ERROR\n");
                        return;
                    }
                    break;
                }
                case '-':{
                    cur--;
                    if(abs(cur)>up_bound){
                        printf("OVERFLOW ERROR\n");
                        return;
                    }
                    break;
                }
                case '@':{
                    if(cur){ p.dir = (p.dir+1)%4; }
                    else { p.dir = (p.dir+3)%4; }
                    break;
                }
            }
        step++;
        if(step>TimeLim) {
            printf("TIME LIMIT EXCEEDED\n");
            return;
        }
        p.Move();
        if(p.r<0||p.r>=h||p.c<0||p.c>=w) {
            printf("RUNTIME ERROR\n"); return;
        }
        op = p.read();
    }
}

int main()
{
    scanf("%d%d",&h,&w);
    for(int i = 0; i < h; i++)
        scanf("%s",G[i]);
    scanf("%d",&readLim);
    for(int i = 0; i < readLim; i++){
        scanf("%d",readList+i);
    }
    execute();

    return 0;
}

 

posted @ 2015-07-29 19:37  陈瑞宇  阅读(220)  评论(0编辑  收藏  举报