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 = ;
char dictionary[][WORDLEN];
char Word[][];
int dictSign[]={};
int dictSize;
int a[];
 
static void
readDictionary(char *dictName)
{
    FILE *h;
    char s[];
    int i;
 
    dictSize = ;
    h = fopen(dictName, "rt");
    fgets(s, , h);
    while (!feof(h)) {
        s[strlen(s)-] = '\0';  /* get rid of newline */
            strcpy(dictionary[dictSize++], s);
        fgets(s, , h);
    }
    printf("Dictionary size is %d\n", dictSize);
 
    for(i=; i<dictSize; i++){
        printf("%s\n",dictionary[i]);
    }
}
 
int getMax(){
    int max=,which=;
    int all=;
    for(int i=; i<dictSize; i++){
        if(dictSign[i] == )
            continue;
        int len = strlen(dictionary[i]);
        if(max < len){
            max = len;
            which = i;
            all = ;
            dictSign[i] = ;
        }
    }
    if(all == )
        return -;
    return which;
}
 
bool input(int x,int y,char *s,int direction){
    int len = strlen(s);
 
    switch(direction){
    case UP :
        for(int i=; i<len; i++){
            if(s[i] != Word[x-i][y] && Word[x-i][y] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x-i][y]=s[i];
        break;
    case UP_R:
        for(int i=; i<len; i++){
            if(s[i] != Word[x-i][y+i] && Word[x-i][y+i] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x-i][y+i]=s[i];
        break;
    case RIGHT:
        for(int i=; i<len; i++){
            if(s[i] != Word[x][y+i] && Word[x][y+i] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x][y+i]=s[i];
        break;
    case DOWN_R:
        for(int i=; i<len; i++){
            if(s[i] != Word[x+i][y+i] && Word[x+i][y+i] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x+i][y+i]=s[i];
        break;
    case DOWN:
        for(int i=; i<len; i++){
            if(s[i] != Word[x+i][y] && Word[x+i][y] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x+i][y]=s[i];
        break;
    case DOWN_L:
        for(int i=; i<len; i++){
            if(s[i] != Word[x+i][y-i] && Word[x+i][y-i] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x+i][y-i]=s[i];
        break;
    case LEFT:
        for(int i=; i<len; i++){
            if(s[i] != Word[x][y-i] && Word[x][y-i] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x][y-i]=s[i];
        break;
    case UP_L:
        for(int i=; i<len; i++){
            if(s[i] != Word[x-i][y-i] && Word[x-i][y-i] != )
                return false;
        }
        for(int i=; i<len; i++)
            Word[x-i][y-i]=s[i];
        break;
    }
    return true;
}
 
void getWhere(int n){
    int layer;
    int n8=;
    for(int i=; ;i++){
        n8 += i*;
        if(n <= n8){
            layer = i;
            n8 -= i*;
            n = n-n8;
            int many = (n-)/(i*);
 
            switch(many){
                case :
                    a[] = MID-i-+n;
                    a[] = MID-i;
                    break;
                case :
                    a[] = MID-i+(n-)%(i*);
                    a[] = MID+i;
                    break;
                case :
                    a[] = MID+i-(n-)%(i*);
                    a[] = MID+i;
                    break;
                case :
                    a[] = MID+i-(n-)%(i*);
                    a[] = 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[];
    input(MID,MID,s,DOWN);
    //printf("%s\n",s);
 
    for(int i=,j=; i<dictSize-;i++){
        int judge=;
        s = dictionary[i+];
 
        if(debug){
            printf("%d %s\n",i,s);
        }
        for(; ;j++){
            getWhere(j);
            int startD = (getDirect(a[],a[])+)%;
            for(int k=; k<=; k++){
                if(input(a[],a[],s,startD%)){
                    judge = ;
                    break;
                }
            }
            if(judge == )
                break;
        }
    }
}
 
int main(int argc, char **argv)
{
    //printf("sb%s%d\n",*argv,argc);
    /* read in the dictionary */
    readDictionary(argv[]);
    ws_maker();
    //input(8,2,"love",UP);
    //input(8,1,"fove",UP_R);
    if(debug)
    for(int i=; i<; i++){
        for(int j=; j<; 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