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URAL 2027 URCAPL, Episode 1 (模拟)

2015-07-29 19:37 405 查看
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;
}
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