poj 1035（水题，注意增删时的操作）

Posted 2021-11-11 智人心

#include<iostream>
#include<cstring>
using namespace std;
int main(){
    char dict[10005][20],tar[20],tmp[20];
    int count,i,j,diff_num;
    bool flag;
    count = 0;
    while(scanf("%s",tmp)==1&&strcmp(tmp,"#")!=0){
        strcpy(dict[count++],tmp);
    }
    while(scanf("%s",tmp)==1&&strcmp(tmp,"#")!=0){
        flag = false;
        for(i=0;i<count;i++){
            if(strcmp(tmp,dict[i])==0){
                flag = true;
            }
        }
        if(flag){
            printf("%s is correct\\n",tmp);
            continue;
        }
        printf("%s:",tmp);
        for(i=0;i<count;i++){
            if(strlen(tmp)==strlen(dict[i])){
                diff_num = 0;
                for(j=0;j<strlen(tmp);j++){
                    if(tmp[j]!=dict[i][j])
                        diff_num++;
                }
                if(diff_num==1)
                    printf(" %s",dict[i]);
            }
            else if(strlen(tmp)==strlen(dict[i])+1){
                j = 0;
                while(j<strlen(dict[i])&&tmp[j]==dict[i][j])j++;
                while(j<strlen(dict[i])&&tmp[j+1]==dict[i][j])j++;
                if(j==strlen(dict[i]))
                    printf(" %s",dict[i]);
            }
            else if(strlen(tmp)==strlen(dict[i])-1){
                j = 0;
                while(j<strlen(tmp)&&tmp[j]==dict[i][j])j++;
                while(j<strlen(tmp)&&tmp[j]==dict[i][j+1])j++;
                if(j==strlen(tmp))
                    printf(" %s",dict[i]);
            }
        }
        printf("\\n"); 
    }
    return 0;
} 

 

POJ题目总结

 (1)深度优先搜索 (poj2488,poj3083,poj3009,poj1321,poj2251)
(2)广度优先搜索(poj3278,poj1426,poj3126,poj3087.poj3414)
(3)简单搜索技巧和剪枝(poj2531,poj1416,poj2676,1129)
(1)枚举. (poj1753,poj2965)
(2)贪心(poj1328,poj2109,poj2586)
(6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)

简单模拟
1006, 1008, 1013, 1016, 1017, 1169, 1298, 1326, 1350, 1363, 1676, 1786,
1791, 1835, 1970, 2317, 2325, 2390,