codeforces 156 div1

最新推荐文章于 2025-01-20 09:00:00 发布

ls_0222

最新推荐文章于 2025-01-20 09:00:00 发布

阅读量1k

点赞数

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/ls_0222/article/details/8333551

版权

ACM 专栏收录该内容

9 篇文章

订阅专栏

本文解析了两道经典的编程题目，第一题是如何寻找由交替出现的两个数组成的子串的最大长度；第二题则探讨了如何通过二分搜索确定在二维网格中遍历指定数量节点所需的最小步数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A：给你n个数，求由交替出现的两个数组成的子串的最大长度

先离散化，dp[i][j]表示当前以i结尾，下一个需要j的最大长度

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <set>

#define pause cout << " press ansy key to continue...",  cin >> chh
#define file_r(x) freopen(x,  "r",  stdin)
#define file_w(x) freopen(x,  "w",  stdout)
#define lowbit(x) ((x) & (-x))
#define repit(i, c) for (__typeof__((c).begin()) i = (c).begin(); i != (c).end(); i++)
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repd(i, n) for (int i = (n - 1); i >= 0; i--)
#define FOR(i, n, m) for (int i = (n); i <= (m); i++)
#define FORD(i, n, m) for (int i = (n); i >= (m); i--)
#define mem(a, x) memset(a, x, sizeof(a))
#define pb push_back
#define X first
#define Y second
#define mp make_pair
#define sqr(r) ((r) * (r))
#define dis(x1, y1, x2, y2) (((x1) - (x2)) * ((x1) - (x2)) + ((y1) - (y2)) * ((y1) - (y2)))

#define debug1(x) cout << #x" = " << x << endl
#define debug2(x, y) cout << #x" = " << x << " " << #y" = " << y << endl
#define debug3(x, y, z) cout << #x" = " << x << " " << #y" = " << y << " " << #z" = " << z << endl
#define debug4(x, y, z, w) cout << #x" = " << x << " " << #y" = " << y << " " << #z" = " << z << " " << #w" = " << w << endl

using namespace std;
int chh;

typedef vector<int> vi;
typedef set<int> si;
typedef map<int, int> mii;
typedef pair<int, int> pii;
typedef long long LL;

const int N = 4005;

int n, e;
int dp[N][N];
int a[N], b[N];

int main() {
    int u, ans;
    while (~scanf("%d", &n)) {
        FOR (i, 1, n) {
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        if (n <= 2) {
            printf("%d\n", n);
            continue;
        }
        sort(b + 1, b + n + 1);
        e = unique(b + 1, b + n + 1) - b;
        FOR (i, 1, n) a[i] = lower_bound(b + 1, b + e, a[i]) - b;
        memset(dp, -1, sizeof(dp));
        dp[a[1]][e + 1] = 1;
        FOR (i, 2, n) {
            u = a[i];
            FOR (j, 1, e) {
                if (dp[j][u] != -1) dp[u][j] = max(dp[u][j], dp[j][u] + 1);
                if (dp[j][e + 1] != -1) {
                    dp[u][j] = max(dp[u][j], 2);
                }
            }
            dp[u][e + 1] = 1;
        }
        ans = 2;
        FOR (i, 1, e) FOR (j, 1, e) ans = max(ans, dp[i][j]);
        printf("%d\n", ans);
    }
    return 0;
}

B：在n * n的地图中给你一个点，每次可以向四个方向扩展，问最少扩展多少次后，

遍历的点大于等于c

二分枚举答案

int n, x, y, c;

LL doIt(int l, int u, int s, int x, int y) {
    int h = max(u, x - (s - y + l));
    LL ans = (LL)(x - h + 1) * (y - l + 1);
    if (h - 1 >= u) ans += (LL)(u + 2 * y - 2 * l - h + 1) * (h - u) / 2;
    return ans;
}

LL cal(int s) {
    int h;
    int l = max(1, y - s), r = min(n, y + s);
    int u = max(1, x - s), d = min(n, x + s);
    LL ans = 0;
    ans += doIt(l, u, s, x, y);
    ans += doIt(y - (r - y), u, s, x, y);
    ans += doIt(l, x - (d - x), s, x, y);
    ans += doIt(y - (r - y), x - (d - x), s, x, y);
    ans -= r - l + 1 + d - u + 1 + 1;
    return ans;
}

int main() {
    int s, t, mid;
    LL tol;
    while (~scanf("%d %d %d %d", &n, &x, &y, &c)) {
        if (c == 1) {
            puts("0");
            continue;
        }
        s = 1, t = 1000000000;
        while (s != t) {
            mid = (s + t) >> 1;
            tol = cal(mid);
            if (tol >= c) t = mid;
            else s = mid + 1;
        }
        printf("%d\n", s);
    }
    return 0;
}