POJ 1068&&2632&&1573&&2993&&2996

　　这次的题目是著名的模拟（水逼）专题

　　题目难度都不大，思维也不深，就是一个字——烦

　　同时很多题目都有很多坑点

　　1608 题意是告诉你一个只有（）的字符串（且匹配正确），每个）的左边有几个（，让你求每一对括号中包含了几对括号。

　　思路很水，模拟出原字符串再开个栈做一次就行了。

　　CODE

#include<cstdio>
using namespace std;
const int N=50;
int t,n,s[N],stack[N],i,j,x,last,del,top;
inline void read(int &x)
{
    x=0; char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
}
inline void work(int b,int e)
{
    int l=0,r=0;
    for (int i=b;i<=e;++i)
    if (!s[i]) l++; else r++;
    printf("%d ",l<r?l:r);
}
int main()
{
    read(t);
    while (t--)
    {
        read(n); last=1; del=0;
        for (i=1;i<=n;++i)
        {
            read(x);
            for (j=last;j<last+x-del;++j)
            s[j]=0; 
            s[last+x-del]=1; last=last+x-del+1; del=x; 
        }
        for (i=1;i<=2*n;++i)
        if (!s[i]) stack[++top]=i; else work(stack[top--],i);
        putchar('\n');
    }
    return 0;
}

 

　　2632 就是一堆机器人行走让你求它们会不会撞到或者撞墙。模拟整个过程即可（毫无算法可言）

　　CODE

#include<cstdio>
using namespace std;
const int N=105;
struct robot
{
    int x,y;
    char side;
}r[N];
struct place
{
    bool b;
    int num;
}f[N][N];
int t,a,b,m,n,i,j,x,num,res;
char ch;
bool flag;
inline void read(int &x)
{
    x=0; char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
}
inline void turn_left(int num,int step)
{
    if (r[num].side=='N') res=0; 
    if (r[num].side=='W') res=1; 
    if (r[num].side=='S') res=2; 
    if (r[num].side=='E') res=3; 
    res=(res+step)%4;
    if (res==0) r[num].side='N'; 
    if (res==1) r[num].side='W';
    if (res==2) r[num].side='S';
    if (res==3) r[num].side='E';
}
inline void turn_right(int num,int step)
{
    if (r[num].side=='N') res=0; 
    if (r[num].side=='E') res=1; 
    if (r[num].side=='S') res=2; 
    if (r[num].side=='W') res=3; 
    res=(res+step)%4;
    if (res==0) r[num].side='N'; 
    if (res==1) r[num].side='E';
    if (res==2) r[num].side='S';
    if (res==3) r[num].side='W';
}
inline void forward(int num,int step)
{
    if (!step) return;
    f[r[num].x][r[num].y].b=f[r[num].x][r[num].y].num=0;
    if (r[num].side=='N')
    {
        while (step--)
        {
            if (f[r[num].x][++r[num].y].b) { printf("Robot %d crashes into robot %d\n",num,f[r[num].x][r[num].y].num); flag=1; return; }
            if (r[num].y>b) { printf("Robot %d crashes into the wall\n",num); flag=1; return; }
        }
    }
    if (r[num].side=='E')
    {
        while (step--)
        {
            if (f[++r[num].x][r[num].y].b) { printf("Robot %d crashes into robot %d\n",num,f[r[num].x][r[num].y].num); flag=1; return; }
            if (r[num].x>a) { printf("Robot %d crashes into the wall\n",num); flag=1; return; }
        }
    }
    if (r[num].side=='S')
    {
        while (step--)
        {
            if (f[r[num].x][--r[num].y].b) { printf("Robot %d crashes into robot %d\n",num,f[r[num].x][r[num].y].num); flag=1; return; }
            if (!r[num].y) { printf("Robot %d crashes into the wall\n",num); flag=1; return; }
        }
    }
    if (r[num].side=='W')
    {
        while (step--)
        {
            if (f[--r[num].x][r[num].y].b) { printf("Robot %d crashes into robot %d\n",num,f[r[num].x][r[num].y].num); flag=1; return; }
            if (!r[num].x) { printf("Robot %d crashes into the wall\n",num); flag=1; return; }
        }
    }
    f[r[num].x][r[num].y].b=1; f[r[num].x][r[num].y].num=num;
}
int main()
{
    read(t);
    while (t--)
    {
        for (i=0;i<N;++i)
        for (j=0;j<N;++j)
        f[i][j].b=f[i][j].num=0; flag=0;
        read(a); read(b);
        read(n); read(m);
        for (i=1;i<=n;++i)
        read(r[i].x),read(r[i].y),r[i].side=getchar(),f[r[i].x][r[i].y].b=1,f[r[i].x][r[i].y].num=i;
        while (m--)
        {
            read(num); ch=getchar(); read(x);
            if (flag) continue;
            if (ch=='L') turn_left(num,x);
            if (ch=='R') turn_right(num,x);
            if (ch=='F') forward(num,x);
        }
        if (!flag) puts("OK");
    }
    return 0;
}

 

　　1573 同上，一个机器人在地图上走。让你求要走多久才能走出去或是在循环的圈长。

　　DFS或模拟都可以，难度也不大

　　CODE

#include<cstdio>
#include<cstring>
using namespace std;
const int N=20;
char a[N][N];
int f[N][N],n,m,x,y,i,j;
inline void read(int &x)
{
    x=0; char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
}
int main()
{
    for (;;)
    {
        read(n); read(m); read(y); x=1;
        if (n+m+y==0) break;
        memset(f,0,sizeof(f));
        for (i=1;i<=n;++i)
        {
            for (j=1;j<=m;++j)
            a[i][j]=getchar();
            getchar();
        }
        f[x][y]=1;
        for (;;)
        {
            if (a[x][y]=='N') { if (x==1) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x-1][y]) f[x-1][y]=f[x][y]+1,x--; else { printf("%d step(s) before a loop of %d step(s)\n",f[x-1][y]-1,f[x][y]-f[x-1][y]+1); break; } }
            if (a[x][y]=='E') { if (y==m) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x][y+1]) f[x][y+1]=f[x][y]+1,y++; else { printf("%d step(s) before a loop of %d step(s)\n",f[x][y+1]-1,f[x][y]-f[x][y+1]+1); break; } }
            if (a[x][y]=='S') { if (x==n) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x+1][y]) f[x+1][y]=f[x][y]+1,x++; else { printf("%d step(s) before a loop of %d step(s)\n",f[x+1][y]-1,f[x][y]-f[x+1][y]+1); break; } }
            if (a[x][y]=='W') { if (y==1) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x][y-1]) f[x][y-1]=f[x][y]+1,y--; else { printf("%d step(s) before a loop of %d step(s)\n",f[x][y-1]-1,f[x][y]-f[x][y-1]+1); break; } }
        }
    }
    return 0;
}

 

　　2993&&2996 两道题刚好相反，是一个模拟棋盘的问题。

　　但主要是有一些没有任何用处的东西（+，-，|）影响了正常的操作。

　　注意下细节以及题意，多打几次就A了

　　2993CODE

#include<cstdio>
#include<iostream>
#include<string>
using namespace std;
char a[17][33]=
{
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},
    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},
};
string s1,s2;
inline void white(string s)
{
    while (s.size())
    {
        string temp="";
        for (int i=0;i<=s.find(",");++i)
        temp+=s[i];
        if (temp[0]>='A'&&temp[0]<='Z') a[(8-temp[2]+'0'+1)*2-1][(temp[1]-'a'+1)*4-2]=temp[0],s.erase(0,4); else a[(8-temp[1]+'0'+1)*2-1][(temp[0]-'a'+1)*4-2]='P',s.erase(0,3);
    }
}
inline void black(string s)
{
    while (s.size())
    {
        string temp="";
        for (int i=0;i<=s.find(",");++i)
        temp+=s[i];
        if (temp[0]>='A'&&temp[0]<='Z') a[(8-temp[2]+'0'+1)*2-1][(temp[1]-'a'+1)*4-2]=temp[0]-'A'+'a',s.erase(0,4); else a[(8-temp[1]+'0'+1)*2-1][(temp[0]-'a'+1)*4-2]='p',s.erase(0,3);
    }
}
int main()
{    
    cin>>s1; cin>>s2; s2+=',';
    if (s1[0]=='W') white(s2); else black(s2);
    cin>>s1; cin>>s2; s2+=',';
    if (s1[0]=='W') white(s2); else black(s2);
    for (int i=0;i<17;++i)
    {
        for (int j=0;j<33;++j)
        putchar(a[i][j]);
        putchar('\n');
    }
    return 0;
}

 

　　2996CODE

#include<cstdio>
#include<iostream>
#include<string>
using namespace std;
string a[10],s;
int i,j;
int main()
{
    cin>>s;
    for (i=1;i<=8;++i)
    cin>>a[i],cin>>s;
    printf("White: ");
    for (i=8;i;--i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='K') printf("K%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=8;i;--i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='Q') printf(",Q%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=8;i;--i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='R') 
    printf(",R%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=8;i;--i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='B') 
    printf(",B%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=8;i;--i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='N') printf(",N%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=8;i;--i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='P') 
    printf(",%c%d",'a'+(j+2)/4-1,8-i+1);
    printf("\nBlack: ");
    for (i=1;i<=8;++i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='k') printf("K%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=1;i<=8;++i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='q') printf(",Q%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=1;i<=8;++i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='r') 
    printf(",R%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=1;i<=8;++i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='b') 
    printf(",B%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=1;i<=8;++i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='n') printf(",N%c%d",'a'+(j+2)/4-1,8-i+1);
    for (i=1;i<=8;++i)
    for (j=0;j<a[i].size();++j)
    if (a[i][j]=='p') 
    printf(",%c%d",'a'+(j+2)/4-1,8-i+1);
    return 0;
}

 

　　最后发现我模拟题的代码都是复制来复制去的，导致程序看起来很。。。（臃肿）

　　下次还是尽量多清晰的分块吧