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  • POJ1035 串

    Sample Input

    i
    is
    has
    have
    be
    my
    more
    contest
    me
    too
    if
    award
    #
    me
    aware
    m
    contest
    hav
    oo
    or
    i
    fi
    mre
    #

    Sample Output

    me is correct
    aware: award
    m: i my me
    contest is correct
    hav: has have
    oo: too
    or:
    i is correct
    fi: i
    mre: more me

    这道题首先看到题目数据范围1e4,所以可以加剪枝暴力求解。

    重点是要分析出长度相差1的和长度相等的才可以经过变换的到字典中的串。

    在此基础上,查看匹配情况,长串作为被匹配的, 相同的字符ans++。

    分情况分析是否能够变换得到。

    最后注意输出格式和要求。

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cctype>
    #include <iostream>
    #include <algorithm>
    #include <map>
    #include <set>
    #include <vector>
    #include <string>
    #include <stack>
    #include <queue>
    
    typedef long long LL;
    typedef unsigned long long ULL;
    using namespace std;
    
    // bool Sqrt(LL n) { return (LL)sqrt(n) * sqrt(n) == n; }
    const double PI = acos(-1.0), ESP = 1e-10;
    const LL INF = 99999999999999;
    const int inf = 999999999, maxn = 1e4 + 24;
    string m, a;
    // vector<int> len1, len2;
    vector<string> v1, v2;
    
    int main()
    {
        //freopen("in.txt", "r", stdin);
        //freopen("out.txt", "w", stdout);
        while(cin >> m) {
            if(m == "#") break;
            v1.push_back(m); //len1.push_back(m.size());
        }
        while(cin >> a) {
            if(a == "#") break;
            v2.push_back(a); //len2.push_back(a.size());
        }
        int x = v1.size(), y = v2.size();
        for(int i = 0; i < y; i++) {
            // if(find(v1.begin(), v1.end(), v2[i]) - v1.end() != v1.size() ) cout << v2[i] << " is correct";
            bool flag = 0;
            for(int k = 0; k < x; k++) {
                if(v1[k] == v2[i]) { flag = 1; break;}
            }
            if(flag) { cout << v2[i] << " is correct
    "; continue; }
            else {
                cout << v2[i] << ":";
                int l = v2[i].size();
                for(int j = 0; j < x; j++) {
                    int h = v1[j].size();
                    int ans = 0;
                    if(l == h + 1) {
                        for(int m = 0, d = 0; m < l; m++)
                            if(v1[j][d] == v2[i][m]) { ans++; d++; }
                    }
                    else if(l + 1 == h) {
                        for(int m = 0, d = 0; m < h; m++)
                            if(v1[j][m] == v2[i][d]) { ans++; d++; }
                    }
                    else if(l == h) {
                        for(int m = 0, d = 0; m < l; m++, d++)
                            if(v1[j][d] == v2[i][m]) ans++;
                    }
                    if((l >= h) && (ans == l - 1)) cout << " " << v1[j];
                    if((l < h) && (ans == l)) cout << " " << v1[j];
                }
            }
            puts("");
        }
    
        return 0;
    }
    /*
        input:
        output:
        modeling:
        methods:
        complexity:
        summary:
    */
    View Code
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  • 原文地址:https://www.cnblogs.com/000what/p/11546710.html
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