USACO 之 Section 1.2 Complete Search （已解决）

Milking Cows :  对输入的工作时间做排序处理，之后模拟工作过程，求出答案。。。

/*
ID: Jming
PROG: milk2
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAX_N = 5005;
int N;
struct node {
    int a, b;
};

node Time[MAX_N];

bool Cmp(const node &n1, const node &n2) {
    if (n1.a == n2.a) return n1.b < n2.b;
    else return n1.a < n2.a;
}

int main()
{
    freopen("milk2.in", "r", stdin);
        freopen("milk2.out", "w", stdout);
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%d %d", &Time[i].a, &Time[i].b);
    }
    sort(Time, Time+N, Cmp);
    int mybegin = Time[0].a, myend = Time[0].b;
    int work = myend - mybegin, no_work = 0;
    for (int i = 1; i < N; ++i) {
        if (Time[i].a <= myend) {
            myend = max(myend, Time[i].b);
        }else {
            work = max(work, myend - mybegin);
            no_work = max(no_work, Time[i].a - myend);
            mybegin = Time[i].a, myend = Time[i].b;
        }
    }
    work = max(work, myend - mybegin);
    printf("%d %d\n", work, no_work);
    return 0;
}

 

Transformations ：将转换方法在纸上模拟一下，可发现规律，之后变成程序处理即可。。。

/*
ID: Jming
PROG: transform
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX_N = 15;
char arr_str[MAX_N][MAX_N], t_arr_str[MAX_N][MAX_N], to_arr_str[MAX_N][MAX_N];
int N;

bool Judge(char str1[][15], char str2[][15]) {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            if (str1[i][j] != str2[i][j]) {
                return false;
            }
        }
    }
    return true;
}

void number1(char str1[][15], char str2[][15]) {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            str1[j][N - 1 - i] = str2[i][j];
        }
    }
}

void number2(char str1[][15], char str2[][15]) {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            str1[N - 1 - i][N - 1 - j] = str2[i][j];
        }
    }
}

void number3(char str1[][15], char str2[][15]) {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            str1[N - 1 - j][i] = str2[i][j];
        }
    }
}

void number4(char str1[][15], char str2[][15]) {
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            str1[i][N - 1 - j] = str2[i][j];
        }
    }
}

void Solve() {
    number1(t_arr_str, arr_str);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("1\n");
        return;
    }

    number2(t_arr_str, arr_str);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("2\n");
        return;
    }

    number3(t_arr_str, arr_str);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("3\n");
        return;
    }


    number4(t_arr_str, arr_str);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("4\n");
        return;
    }


    char arr_str1[MAX_N][MAX_N];
    for (int i = 0; i < N; ++i) {
        strcpy(arr_str1[i], t_arr_str[i]);
    }

    number1(t_arr_str, arr_str1);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("5\n");
        return;
    }

    number2(t_arr_str, arr_str1);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("5\n");
        return;
    }

    number3(t_arr_str, arr_str1);
    if (Judge(t_arr_str, to_arr_str)) {
        printf("5\n");
        return;
    }
    if (Judge(t_arr_str, to_arr_str)) {
        printf("6\n");
        return;
    }
    printf("7\n");
}

int main()
{
    freopen("transform.in", "r", stdin);
    freopen("transform.out", "w", stdout);
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%s", arr_str[i]);
        strcpy(t_arr_str[i], arr_str[i]);
    }
    for (int i = 0; i < N; ++i) {
        scanf("%s", to_arr_str[i]);
    }
    Solve();
    return 0;
}

 

Name That Number ：字典树 + Dfs  （解法同  hdu 1298  解题报告点击此处）

/*
ID: Jming
PROG: namenum
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <fstream>
#include <string>
#include <cstring>
#include <map>
using namespace std;
const int MAX = 26;
map<int, string> myMap;
bool ansJudge;

struct Trie{
    bool Judge;
    Trie* next[MAX];
    Trie() {
        Judge = false;
        memset(next, NULL, sizeof(next));
    }
};

Trie* Root;

void CreTrie(string str) {
    int len = str.size();
    Trie* p = Root;
    for (int i = 0; i < len; ++i) {
        int pos = str[i] - 'A';
        if (!(p->next[pos])) {
            p->next[pos] = new Trie;
        }
        p = p->next[pos];
    }
    p->Judge = true;
}

void DelTrie(Trie* T) {
    for (int i = 0; i < MAX; ++i) {
        if (T->next[i]) {
            DelTrie(T->next[i]);
        }
    }
    delete[] T;
}

void Ini() {
    ifstream fcin("dict.txt");
    if (!fcin) {
        system("pause");
    }
    string str;
    while (fcin >> str) {
        CreTrie(str);
    }
    myMap[2] = "ABC"; myMap[3] = "DEF"; myMap[4] = "GHI"; myMap[5] = "JKL";
    myMap[6] = "MNO"; myMap[7] = "PRS"; myMap[8] = "TUV"; myMap[9] = "WXY";
}

void Dfs(string str, int pos, Trie* T, string ansStr) {
    for (int i = 0; i < 3; ++i) {
        Trie* t_Trie = T->next[myMap[str[pos] - '0'][i] - 'A'];
        if (t_Trie) {
            if ((t_Trie->Judge) && (pos == (str.size() - 1))) {
                if (!ansJudge) ansJudge = true;
                cout << ansStr + myMap[str[pos] - '0'][i] << endl;
            }
            else Dfs(str, pos + 1, t_Trie, ansStr + myMap[str[pos] - '0'][i]);
        }
    }
}

int main()
{
   Root = new Trie;
   ansJudge = false;
    Ini();
    freopen("namenum.in", "r", stdin);
    freopen("namenum.out", "w", stdout);
    string str;
    cin >> str;
    Dfs(str, 0, Root, "");
    if (!ansJudge) {
        cout << "NONE" << endl;
    }
    DelTrie(Root);
    return 0;
}

 

Palindromic Squares ：进制处理。。。

/*
ID: Jming
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
using namespace std;
int base;

struct myNode {
    int number;
    int square;
}node[305];

void Init() {
    for (int i = 1; i <= 300; ++i) {
        node[i - 1].number = i;
        node[i - 1].square = i * i;
    }
}

char judgeChar(int tmp) {
    if (tmp >= 0 && tmp <= 9) return ('0' + tmp);
    else return ('A' + (tmp - 10));
}

string getNum(int x) {
    string str = "";
    while (x >= base) {
        int tmp = x%base;
        x /= base;
        str = judgeChar(tmp) + str;
    }
    return (judgeChar(x) + str);
}

bool judgePalindromic(string str) {
    int myBegin = 0, myEnd = str.size() - 1, myCount = str.size() >> 1;
    for (int i = 0; i < myCount; ++i) {
        if (str[myBegin + i] != str[myEnd - i]) {
            return false;
        }
    }
    return true;
}

void Solve() {
    for (int i = 0; i < 300; ++i) {
        string str = getNum(node[i].square);
        if (judgePalindromic(str)) {
            cout << (getNum(node[i].number)) << " " << str << endl;
        }
    }
}

int main()
{
    freopen("palsquare.in", "r", stdin);
    freopen("palsquare.out", "w", stdout);
    Init();
    scanf("%d", &base);
    Solve();
    return 0;
}

 

Dual Palindromes ：进制处理。。。

/*
ID: Jming
PROG: dualpal
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
using namespace std;
int base, N, S;

string getNum(int x) {
    string str = "";
    while (x >= base) {
        int tmp = x%base;
        x /= base;
        str = (char)('0' + tmp) + str;
    }
    return ((char)('0' + x) + str);
}

bool judgePalindromic(string str) {
    int myBegin = 0, myEnd = str.size() - 1, myCount = str.size() >> 1;
    for (int i = 0; i < myCount; ++i) {
        if (str[myBegin + i] != str[myEnd - i]) {
            return false;
        }
    }
    return true;
}

void Solve() {
    int myCount = 1;
    while (myCount <= N) {
        ++S;
        int sum = 0;
        for (base = 2; base <= 10; ++base) {
            string str = getNum(S);
            if (judgePalindromic(str)) {
                ++sum;
                if (sum >= 2) break;
            }
        }
        if (sum >= 2) {
            printf("%d\n", S);
            ++myCount;
        }
    }
}

int main()
{
    freopen("dualpal.in", "r", stdin);
    freopen("dualpal.out", "w", stdout);
    scanf("%d %d", &N, &S);
    Solve();
    return 0;
}