homework-04 抓瞎

程序截图

这是基本版本截图。。。。空都没填上

四个角的话 可以留下四个单词 最后添上就行

 

 

 程序思路

在一个大平面上先一个中心，然后从中心向四周填词

每次填词时，寻找一个使得矩阵最小的

代目如下

#include <stdio.h>

#include <string.h>

#include <ctype.h>



#define WORDLEN 12

#define MID 15

#define UP 0

#define UP_R 1

#define RIGHT 2

#define DOWN_R 3

#define DOWN 4

#define DOWN_L 5

#define LEFT 6

#define UP_L 7



int debug = 1;

char dictionary[100][WORDLEN];

char Word[100][100];

int dictSign[100]={0};

int dictSize;

int a[2];



static void

readDictionary(char *dictName)

{

    FILE *h;

    char s[80];

    int i;



    dictSize = 0;

    h = fopen(dictName, "rt");

    fgets(s, 80, h);

    while (!feof(h)) {

        s[strlen(s)-1] = '\0';  /* get rid of newline */

            strcpy(dictionary[dictSize++], s);

        fgets(s, 80, h);

    }

    printf("Dictionary size is %d\n", dictSize);



    for(i=0; i<dictSize; i++){

        printf("%s\n",dictionary[i]);

    }

}



int getMax(){

    int max=0,which=0;

    int all=0;

    for(int i=0; i<dictSize; i++){

        if(dictSign[i] == 1)

            continue;

        int len = strlen(dictionary[i]);

        if(max < len){

            max = len;

            which = i;

            all = 1;

            dictSign[i] = 1;

        }

    }

    if(all == 0)

        return -1;

    return which;

}



bool input(int x,int y,char *s,int direction){

    int len = strlen(s);



    switch(direction){

    case UP :

        for(int i=0; i<len; i++){

            if(s[i] != Word[x-i][y] && Word[x-i][y] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x-i][y]=s[i];

        break;

    case UP_R:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x-i][y+i] && Word[x-i][y+i] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x-i][y+i]=s[i];

        break;

    case RIGHT:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x][y+i] && Word[x][y+i] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x][y+i]=s[i];

        break;

    case DOWN_R:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x+i][y+i] && Word[x+i][y+i] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x+i][y+i]=s[i];

        break;

    case DOWN:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x+i][y] && Word[x+i][y] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x+i][y]=s[i];

        break;

    case DOWN_L:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x+i][y-i] && Word[x+i][y-i] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x+i][y-i]=s[i];

        break;

    case LEFT:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x][y-i] && Word[x][y-i] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x][y-i]=s[i];

        break;

    case UP_L:

        for(int i=0; i<len; i++){

            if(s[i] != Word[x-i][y-i] && Word[x-i][y-i] != 0)

                return false;

        }

        for(int i=0; i<len; i++)

            Word[x-i][y-i]=s[i];

        break;

    }

    return true;

}



void getWhere(int n){

    int layer;

    int n8=0;

    for(int i=1; ;i++){

        n8 += i*8;

        if(n <= n8){

            layer = i;

            n8 -= i*8;

            n = n-n8;

            int many = (n-1)/(i*2);



            switch(many){

                case 0:

                    a[0] = MID-i-1+n;

                    a[1] = MID-i;

                    break;

                case 1:

                    a[1] = MID-i+(n-1)%(i*2);

                    a[0] = MID+i;

                    break;

                case 2:

                    a[0] = MID+i-(n-1)%(i*2);

                    a[1] = MID+i;

                    break;

                case 3:

                    a[1] = MID+i-(n-1)%(i*2);

                    a[0] = MID-i;

                    break;

            }

            break;

        }

    }

}

int getDirect(int x, int y){

    if(x>MID){

        if(y>MID)

            return DOWN_L;

        else if(y==MID)

            return LEFT;

        else

            return UP_L;

    }

    else if(x == MID){

        if(y > MID)

            return DOWN;

        else

            return UP;

    }

    else{

        if(y>MID)

            return DOWN_R;

        else if(y==MID)

            return RIGHT;

        else

            return UP_R;

    }

}



void ws_maker(){

    //printf("%d\n",getMax());

    char *s = dictionary[0];

    input(MID,MID,s,DOWN);

    //printf("%s\n",s);

    

    for(int i=0,j=1; i<dictSize-1;i++){

        int judge=0;

        s = dictionary[i+1];



        if(debug){

            printf("%d %s\n",i,s);

        }

        for(; ;j++){

            getWhere(j);

            int startD = (getDirect(a[0],a[1])+1)%8;

            for(int k=1; k<=8; k++){

                if(input(a[0],a[1],s,startD%8)){

                    judge = 1;

                    break;

                }

            }

            if(judge == 1)

                break;

        }

    }

}







int main(int argc, char **argv)

{

    //printf("sb%s%d\n",*argv,argc);

    /* read in the dictionary */

    readDictionary(argv[1]);

    ws_maker();

    //input(8,2,"love",UP);

    //input(8,1,"fove",UP_R);

    if(debug)

    for(int i=0; i<30; i++){

        for(int j=0; j<30; j++)

            printf("%C",Word[i][j]);

        printf("\n");

    }



}

 

 

 

	Personal Software Process Stages	时间百分比(%)	实际花费的时间 (分钟)	原来估计的时间 (分钟)
Planning	计划	20	30	0
·         Estimate	·         估计这个任务需要多少时间，把工作细化并大致排序	20	30	0
Development	开发	60	90	70
·         Analysis	·         需求分析 (包括学习新技术)	10	15	0
·         Design Spec	·         生成设计文档	0	0	0
·         Design Review	·         设计复审 (和同事审核设计文档)	0	0	0
·         Coding Standard	·         代码规范 (制定合适的规范)	0	0	0
·         Design	·         具体设计	5	7.5	10
·         Coding	·         具体编码	20	30	45
·         Code Review	·         代码复审	5	7.5	10
·         Test	·         测试（自我测试，修改代码，提交修改）	20	30	5
Reporting	总结报告	20	30	30
·         Test Report	·         测试报告	10	15	10
·         Size Measurement	·         计算工作量	5	7.5	10
·         Postmortem & Improvement Plan	·         事后总结, 并提出改进	5	7.5	10
Total	总计	100%	总用时150	总估计的用时 100