USACO training chapter1 入门/section1.3

第二部分还是暴力为主。

1.Milking Cows

不是很难，对区间的左端点排序，然后遍历一遍。
可能会有只有一个工人的情况，记得处理。

/*
ID: b1703011
TASK: milk2
LANG: C++                 
*/
#include <iostream>
#include <algorithm>
using namespace std;
const int max_n = 5010;
struct T
{
	int l, r;
	bool operator < (const T &t) const {
		return l < t.l;
	}
} F[max_n];

int main() {
	freopen("milk2.in", "r", stdin);
	freopen("milk2.out", "w", stdout);
	int n;
	cin >> n;
	
	for (int i = 0; i < n; i++) 
		cin >> F[i].l >> F[i].r;

	sort(F, F + n);

	int res1 = -1, res2 = 0;
	int lstr = -1, lstl = -1;
	for (int i = 0; i < n; i++) {
		if (F[i].l > lstr) {
			if (lstr != -1) {
				res2 = max(res2, F[i].l - lstr);
			}
			lstl = F[i].l;
			lstr = F[i].r;
            res1 = max(res1, F[i].r - F[i].l);
		}
		else {
			lstr = max(lstr, F[i].r);
			res1 = max(res1, lstr - lstl);
		}
	}

	cout << res1 << " " << res2 << endl;
	return 0;
}

2. Transformations

这个题绝了，神烦。难度不高，就是把一个矩阵的元素分别翻转90度，180度，270度，镜像，然后在镜像后翻转90度，180度，270度。。。。。。。。
写的不是很简洁，没改了。

/*
ID: b1703011
TASK: transform
LANG: C++                 
*/
#include <iostream>
#include <cstring>
using namespace std;
int n;
char s[12][12], t[12][12], res[12][12];

bool check() {
    bool f = true;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if(t[i][j] != res[i][j]) {
                f = false;
                break;
            }
        }
    }
    return f;
}

bool rotate_w1() {
    //90度旋转
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            t[j][n-1-i] = s[i][j];
        }
    }
    return check();
}

bool rotate_w2() {
    //180度旋转
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            t[n-1-i][n-1-j] = s[i][j];
        }
    }
    return check();
}

bool rotate_w3() {
    // 270度旋转
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
             t[n-1-j][i] = s[i][j];
        }
    }
    return check();
}

bool reflect() {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            t[i][n-1-j] = s[i][j]; 
        }
    }    
    return check();
}

bool combination() {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n/2; j++) {
            int tmp = s[i][n-1-j];
            s[i][n-1-j] = s[i][j];
            s[i][j] = tmp;
        }
    }    
    bool f = false;
    if(rotate_w1() || rotate_w2() || rotate_w3()) f = true;
    return f;
}

bool NoChange() {
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            t[i][j] = s[i][j];
        }
    }
    return check();
}

int main() {
    freopen("transform.in", "r", stdin);
    freopen("transform.out", "w", stdout);
    cin >> n;
    for (int i = 0; i < n; i++) cin >> s[i];
    for (int i = 0; i < n; i++) cin >> res[i];
    
    int ans = 7;
    if (NoChange()) ans = 6;
    
    if (rotate_w1()) ans = min(ans, 1);
    else if (rotate_w2()) ans = min(ans, 2);
    else if (rotate_w3()) ans = min(ans, 3);
    else if (reflect()) ans = min(ans, 4);
    else if (combination()) ans = min(ans, 5);
    cout << ans << endl;
    return 0;
}

3.Name That Number

这个题两种解法。
第一种是二分查字典，复杂度较高，需要生成一个排列（递归）。
第二种比较巧妙，每个数字对应多个字母，但是每个字母就对应唯一的数字，直接把字典中每个名字转化成数字，然后和输入对比就好了。

/*
ID: b1703011
TASK: namenum
LANG: C++                 
*/
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const char num[30] = "22233344455566677778889999";

int main() {
    freopen("namenum.in", "r", stdin);
    freopen("namenum.out", "w", stdout);
    char a[20];
    scanf("%s", a);
    int m = strlen(a);
    bool flag = 0;
    freopen("dict.txt", "r", stdin);
    char t[20];
    while (~scanf("%s", t)) {
        int l = strlen(t);
        if (l != m) continue;
        bool f = 1;
        for (int i = 0; i < l; i++) {
            if (num[t[i]-'A'] != a[i]) {
                f = 0; break;
            }
        }
        if (f == 1) {
            printf("%s\n", t);
            flag = 1;
        }
    }
    if (flag == 0) printf("NONE\n");

	return 0;
}

4.Palindromic Squares

这个题我用的string得到b进制数的字符串
，题目很直接，做法也很直接。

/*
ID: b1703011
TASK: palsquare
LANG: C++                 
*/
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int M = 300;

int B;

string intToA(int a) {
    string BNum = "";
    while (a > 0) {
        int y = a % B;
        if (y < 10) BNum += '0' + y;
        else BNum += 'A' + y - 10;
        a /= B;
    }
    reverse(BNum.begin(), BNum.end());
    return BNum;
}

bool check(int a) {
    string BNum = intToA(a);
    int cnt = BNum.length();
    for (int i = 0; i <= cnt/2; i++) {
       if (BNum[i] != BNum[cnt-1-i]) return false;
   }
    return true;
}

int main() {
	freopen("palsquare.in", "r", stdin);
    freopen("palsquare.out", "w", stdout);
    cin >> B;
    string a, b;
    for (int i = 1; i <= M; i++) {
        if (check(i * i)) {
            a = intToA(i);
            b = intToA(i * i);
            cout << a << " " << b << endl;
        }
    }
            
    
    return 0;
}

5. Dual Palindromes

也是很直接。和上题基本一样，换了种写法。

/*
ID: b1703011
TASK: dualpal
LANG: C++                 
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

void getNum(int a, int b, char *s) {
    int i = 0;
    while (a > 0) {
        int t = a % b;
        if (t < 10) s[i] = t+'0';
        else s[i] = t%10+'A';
        a /= b;
        i++;
    }
    for (int j = 0; j < i/2; j++) swap(s[j], s[i-j-1]);
    s[i] = '\0';  // <-------很重要，不写会出错，不知道为什.
}

bool ispal(char *s, int c) {
    for (int i = 0; i < c; i++)
        if (s[i] != s[c-i-1]) return false;
    return true;
}

int main() {
	freopen("dualpal.in", "r", stdin);
	freopen("dualpal.out", "w", stdout);
    int n, s;
    cin >> n >> s;
    for (int i = s + 1; n; i++) {
        int cnt = 0;
        for (int b = 2; b <= 10 && cnt < 2; b++) {
            char ss[20];
            getNum(i, b, ss);
            if (ispal(ss, strlen(ss))) cnt++;
        }
        if (cnt >= 2) {
            cout << i << endl;
            n--;
        }
    }
    
    return 0;
}

预告：下个section是贪心算法（greedy）