Codeforces Round #295 (Div. 2)-优快云博客

本文解析了两道经典的编程题目，第一题通过使用哈希表判断一个字符串是否为全字母句，第二题采用广度优先搜索算法求解从一个数通过*2或-1操作达到另一个数所需的最少步骤。

/*
    水题
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;

int main(void)
{
    //freopen ("A.in", "r", stdin);

    map<char, int> m1;
    map<char, int> m2;
    int n;
    while (~scanf ("%d", &n))
    {
        for (int i=1; i<=26; ++i)
        {
            m1[i] = 0;   m2[i] = 0;
        }
        char s[110];
        scanf ("%s", &s);
        for (int i=0; i<=n-1; ++i)
        {
            if (s[i]<='z' && s[i]>='a')
            {
                int x = s[i] - 'a' + 1;
                m1[x]++;
            }
            else
            {
                int y = s[i] - 'A' + 1;
                m2[y]++;
            }
        }

        bool flag = true;
        for (int i=1; i<=26; ++i)
        {
            if (m1[i] == 0 && m2[i] == 0)
            {
                flag = false;   break;
            }
        }

        (flag) ? puts ("YES") : puts ("NO");
    }

    return 0;
}

BFS B. Two Buttons

题意：给出n，m两个数字，n可以*2，或者-1，问最少几步n变成m
思路：BFS：从n出发，分两条路(a, b)，标记计算后的数字，如果没找到，入队；如果找到了则输出，不入队，BFS结束。详细解释

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <string>
#include <cstring>
using namespace std;

const int MAXN = 2e4 + 10;
const int INF = 0x3f3f3f3f;
int dp[MAXN];
int used[MAXN];
struct NODE
{
    int x;
    int cnt;
}s;

void BFS(int n, int m)
{
    if (n >= m)
    {
        printf ("%d\n", n - m);        return ;
    }
    queue<struct NODE> q;

    s.x = n;    s.cnt = 0;
    used[n] = 1;
    q.push (s);
    
    NODE a, b;
    while (!q.empty())
    {
        a = q.front();    b = q.front();    q.pop();
        a.x *= 2;    a.cnt++;
        b.x -= 1;    b.cnt++;
        if (a.x == m)
        {
            printf ("%d\n", a.cnt);        break;
        }
        if (b.x == m)
        {
            printf ("%d\n", b.cnt);
        }
        if (a.x > 0 && a.x < MAXN && !used[a.x])
        {
            q.push (a);        used[a.x] = 1;
        }
        if (b.x > 0 && b.x < MAXN && !used[b.x])
        {
            q.push (b);        used[b.x] = 1;
        }
    }    
}

int main(void)
{
    //freopen ("B.in", "r", stdin);

    int n, m;
    int ans;
    while (~scanf ("%d%d", &n, &m))
    {
        memset (used, 0, sizeof (used));
        BFS (n, m);
    }
    
    return 0;
}

转载于:https://www.cnblogs.com/Running-Time/p/4366791.html