csu1510 Happy Robot Recursion

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Problem Solving Ideas:

To solve four values X_min,x_max,y_min,y_max

First consider how x_min get: Since the robot finally has the ability to face four directions (e,w,n,s), we can use an array a[4] to save the robot in 4 directions when the value of X

A[0]: x+,a[1]:x-. a[2]:y+,a[3],y-;

Only ask for the a[4] at the end of the minimum value, that is, x_min

The key to seeking a[4] is how to get the optimal solution when running the command:

A[x]=min (turn left, turn right, advance);

namely A[0]=min (a[2],a[3],a[0]+1);

Similarly get a[1],a[2],a[3]

Code:

#include <iostream> #include <cstring> #include <cstdio> #define INF 9999using namespace Std;char str[    1050];int Min (int a,int b,int c) {if (a>b) a=b;    if (a>c) a=c; return A;}    int Max (int a,int b,int c) {if (a<b) a=b;    if (a<c) a=c; return A;}    void Change (int* t,int dir) {int tt=t[0];        if (dir) {t[0]=t[3];        T[3]=T[1];        T[1]=T[2];    T[2]=tt;        } else {t[0]=t[2];        T[2]=T[1];        T[1]=T[3];    T[3]=tt;    }}int Main () {int len;    int t[4];    int case=1;        while (~SCANF ("%s", str)) {int a[4]= {0,-inf,-inf,-inf};        int b[4]= {0,inf,inf,inf};        int c[4]= {0,-inf,-inf,-inf};        int d[4]= {0,inf,inf,inf};        Len=strlen (str);                for (int i=0; i<len; i++) {if (str[i]== ' F ') {a[0]++,a[1]--;                b[0]++,b[1]--;                c[2]++,c[3]--;            d[2]++,d[3]--;    }        else if (str[i]== ' L ') {change (a,1);                Change (b,1);                Change (c,1);            Change (d,1);                } else if (str[i]== ' R ') {change (a,0);                Change (b,0);                Change (c,0);            Change (d,0);                } else {T[0]=max (a[0]+1,a[2],a[3]);                T[1]=max (A[1]-1,a[2],a[3]);                T[2]=max (a[0],a[1],a[2]);                T[3]=max (A[0],a[1],a[3]);                memcpy (a,t,16);                T[0]=min (B[0]+1,b[2],b[3]);                T[1]=min (B[1]-1,b[2],b[3]);                T[2]=min (b[0],b[1],b[2]);                T[3]=min (B[0],b[1],b[3]);                memcpy (b,t,16);                T[0]=max (C[0],c[2],c[3]);                T[1]=max (C[1],c[2],c[3]);                T[2]=max (c[0],c[1],c[2]+1);                T[3]=max (c[0],c[1],c[3]-1);                memcpy (c,t,16); T[0]=min (d[0],d[2],d[3]);                T[1]=min (D[1],d[2],d[3]);                T[2]=min (d[0],d[1],d[2]+1);                T[3]=min (d[0],d[1],d[3]-1);            memcpy (d,t,16);        }} T[0]=max (Max (a[0],a[1],a[2]), a[3]);         T[1]=min (min (b[0],b[1],b[2]), b[3]);        T[2]=max (Max (c[0],c[1],c[2]), c[3]);        T[3]=min (min (d[0],d[1],d[2]), d[3]);    printf ("Case%d:%d%d%d%d\n", case++,t[1],t[0],t[3],t[2]); } return 0;}

csu1510 Happy Robot Recursion