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=;

int t,n,s[N],stack[N],i,j,x,last,del,top;

inline void read(int &x)

{

    x=; char ch=getchar();

    while (ch<''||ch>'') ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

}

inline void work(int b,int e)

{

    int l=,r=;

    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=; del=;

        for (i=;i<=n;++i)

        {

            read(x);

            for (j=last;j<last+x-del;++j)

            s[j]=;

            s[last+x-del]=; last=last+x-del+; del=x;

        }

        for (i=;i<=*n;++i)

        if (!s[i]) stack[++top]=i; else work(stack[top--],i);

        putchar('\n');

    }

    return ;

}

　　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=;

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=; char ch=getchar();

    while (ch<''||ch>'') ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

}

inline void turn_left(int num,int step)

{

    if (r[num].side=='N') res=;

    if (r[num].side=='W') res=;

    if (r[num].side=='S') res=;

    if (r[num].side=='E') res=;

    res=(res+step)%;

    if (res==) r[num].side='N';

    if (res==) r[num].side='W';

    if (res==) r[num].side='S';

    if (res==) r[num].side='E';

}

inline void turn_right(int num,int step)

{

    if (r[num].side=='N') res=;

    if (r[num].side=='E') res=;

    if (r[num].side=='S') res=;

    if (r[num].side=='W') res=;

    res=(res+step)%;

    if (res==) r[num].side='N';

    if (res==) r[num].side='E';

    if (res==) r[num].side='S';

    if (res==) 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=;

    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=; return; }

            if (r[num].y>b) { printf("Robot %d crashes into the wall\n",num); flag=; 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=; return; }

            if (r[num].x>a) { printf("Robot %d crashes into the wall\n",num); flag=; 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=; return; }

            if (!r[num].y) { printf("Robot %d crashes into the wall\n",num); flag=; 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=; return; }

            if (!r[num].x) { printf("Robot %d crashes into the wall\n",num); flag=; return; }

        }

    }

    f[r[num].x][r[num].y].b=; f[r[num].x][r[num].y].num=num;

}

int main()

{

    read(t);

    while (t--)

    {

        for (i=;i<N;++i)

        for (j=;j<N;++j)

        f[i][j].b=f[i][j].num=; flag=;

        read(a); read(b);

        read(n); read(m);

        for (i=;i<=n;++i)

        read(r[i].x),read(r[i].y),r[i].side=getchar(),f[r[i].x][r[i].y].b=,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 ;

}

　　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=;

char a[N][N];

int f[N][N],n,m,x,y,i,j;

inline void read(int &x)

{

    x=; char ch=getchar();

    while (ch<''||ch>'') ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

}

int main()

{

    for (;;)

    {

        read(n); read(m); read(y); x=;

        if (n+m+y==) break;

        memset(f,,sizeof(f));

        for (i=;i<=n;++i)

        {

            for (j=;j<=m;++j)

            a[i][j]=getchar();

            getchar();

        }

        f[x][y]=;

        for (;;)

        {

            if (a[x][y]=='N') { if (x==) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x-][y]) f[x-][y]=f[x][y]+,x--; else { printf("%d step(s) before a loop of %d step(s)\n",f[x-][y]-,f[x][y]-f[x-][y]+); break; } }

            if (a[x][y]=='E') { if (y==m) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x][y+]) f[x][y+]=f[x][y]+,y++; else { printf("%d step(s) before a loop of %d step(s)\n",f[x][y+]-,f[x][y]-f[x][y+]+); break; } }

            if (a[x][y]=='S') { if (x==n) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x+][y]) f[x+][y]=f[x][y]+,x++; else { printf("%d step(s) before a loop of %d step(s)\n",f[x+][y]-,f[x][y]-f[x+][y]+); break; } }

            if (a[x][y]=='W') { if (y==) { printf("%d step(s) to exit\n",f[x][y]); break; } if (!f[x][y-]) f[x][y-]=f[x][y]+,y--; else { printf("%d step(s) before a loop of %d step(s)\n",f[x][y-]-,f[x][y]-f[x][y-]+); break; } }

        }

    }

    return ;

}

　　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[][]=

{

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

    {'|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|',':',':',':','|','.','.','.','|'},

    {'+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+','-','-','-','+'},

};

string s1,s2;

inline void white(string s)

{

    while (s.size())

    {

        string temp="";

        for (int i=;i<=s.find(",");++i)

        temp+=s[i];

        if (temp[]>='A'&&temp[]<='Z') a[(-temp[]+''+)*-][(temp[]-'a'+)*-]=temp[],s.erase(,); else a[(-temp[]+''+)*-][(temp[]-'a'+)*-]='P',s.erase(,);

    }

}

inline void black(string s)

{

    while (s.size())

    {

        string temp="";

        for (int i=;i<=s.find(",");++i)

        temp+=s[i];

        if (temp[]>='A'&&temp[]<='Z') a[(-temp[]+''+)*-][(temp[]-'a'+)*-]=temp[]-'A'+'a',s.erase(,); else a[(-temp[]+''+)*-][(temp[]-'a'+)*-]='p',s.erase(,);

    }

}

int main()

{

    cin>>s1; cin>>s2; s2+=',';

    if (s1[]=='W') white(s2); else black(s2);

    cin>>s1; cin>>s2; s2+=',';

    if (s1[]=='W') white(s2); else black(s2);

    for (int i=;i<;++i)

    {

        for (int j=;j<;++j)

        putchar(a[i][j]);

        putchar('\n');

    }

    return ;

}

　　2996CODE

#include<cstdio>

#include<iostream>

#include<string>

using namespace std;

string a[],s;

int i,j;

int main()

{

    cin>>s;

    for (i=;i<=;++i)

    cin>>a[i],cin>>s;

    printf("White: ");

    for (i=;i;--i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='K') printf("K%c%d",'a'+(j+)/-,-i+);

    for (i=;i;--i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='Q') printf(",Q%c%d",'a'+(j+)/-,-i+);

    for (i=;i;--i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='R')

    printf(",R%c%d",'a'+(j+)/-,-i+);

    for (i=;i;--i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='B')

    printf(",B%c%d",'a'+(j+)/-,-i+);

    for (i=;i;--i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='N') printf(",N%c%d",'a'+(j+)/-,-i+);

    for (i=;i;--i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='P')

    printf(",%c%d",'a'+(j+)/-,-i+);

    printf("\nBlack: ");

    for (i=;i<=;++i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='k') printf("K%c%d",'a'+(j+)/-,-i+);

    for (i=;i<=;++i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='q') printf(",Q%c%d",'a'+(j+)/-,-i+);

    for (i=;i<=;++i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='r')

    printf(",R%c%d",'a'+(j+)/-,-i+);

    for (i=;i<=;++i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='b')

    printf(",B%c%d",'a'+(j+)/-,-i+);

    for (i=;i<=;++i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='n') printf(",N%c%d",'a'+(j+)/-,-i+);

    for (i=;i<=;++i)

    for (j=;j<a[i].size();++j)

    if (a[i][j]=='p')

    printf(",%c%d",'a'+(j+)/-,-i+);

    return ;

}

　　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

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