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天命




离线:45分钟前



天命
4小时前

思路:先将字符串反转,然后以空格为标志,每遇上一个空格,就翻转当前单词。

class Solution {
public:
    string reverseWords(string s) {
        if(s.empty()) return s;

        reverse(s.begin(), s.end());
        int idx = 0, j = 0;//j表示的是这个单词的开始下标, idx表示当前单词的结束下标

        while(idx < s.size()){
            if(s[idx] == ' '){
                reverse(s.begin() + j, s.begin() + idx);
                j = idx + 1;
            }
            idx++;
        }
        reverse(s.begin() + j, s.begin() + s.size());
        return s;
    }
};



天命
14天前
class Solution {
public:
    int findMin(vector<int>& numbers) {
        int n = numbers.size();
        if(!n) return -1;
        int i = 0;
        for(i = 0; i < n; i++){
            if(numbers[i] < numbers[n - 1]) break;
        }
        return numbers[i]; 
    }
};


活动打卡代码 AcWing 1073. 树的中心

天命
2个月前
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 10005, M = N * 2;
int d1[M], d2[M];//每个结点往下走的最大值和次大值,下标存储的是结点的编号
int h[M], e[M], w[M], ne[M], idx;
int p1[M];//p1存储的是该结点的相邻结点
int up[M];//往上走的最远距离
int n;//n个点
const int INF = 0x3F3F3F3F;
int dfs_d(int u, int father)
{
    d1[u] = d2[u] = -INF;//求取max,赋值-INF
    for(int i = h[u]; i != -1; i = ne[i])//遍历所有相邻结点
    {
        int j = e[i];//j是相邻点的编号
        if(j == father) continue;//不走回父亲
        int d = dfs_d(j, u) + w[i];//用子节点来更新父节点
        if(d >= d1[u])
          {
             d2[u] = d1[u];
             d1[u] = d;
             p1[u] = j;//p1用来在dfs_u中判断,下标存储的是它的最大路径走过的第一个结点
          }
        else if(d >= d2[u])
        {
            d2[u] = d;
        }
    }
    if(d1[u] == -INF)//d1[u]是叶子结点,没有更新,赋值为0
        d1[u] = d2[u] = 0;
    return d1[u];
}

void dfs_u(int u, int father)
{
    for(int i = h[u]; i != -1; i = ne[i]){
        int j = e[i];
        if(j == father) continue;//不走回父亲
        if(p1[u] == j)
            up[j] = max(up[u], d2[u]) + w[i];//经过当前点选择次大的路径值
        else up[j] = max(up[u], d1[u]) + w[i];//没经过当前点往下走
        dfs_u(j, u);
    }
}

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int main()
{
    cin >> n;
    memset(h, -1, sizeof h);
    int a, b, c;
    for(int i = 1; i <= n - 1; ++i){
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }

    dfs_d(1, -1);//往下深搜
    dfs_u(1, -1);//往下深搜,但这次遍历当前结点的子结点的时候更新这些子结点的up值,将u当作这些结点往上走的第一个结点

    int res = INF;
    for(int i = 1; i <= n; ++i)
        res = min(res, max(d1[i], up[i]));

    cout << res;


    return 0;
}


活动打卡代码 AcWing 1072. 树的最长路径

天命
2个月前
#include<iostream>
#include<cstring>

using namespace std;
const int N = 10010;

int n;
int h[N], idx, ne[N * 2], e[N * 2], w[N * 2];   //边有来回两条
int ans;

void add(int a, int b, int c){
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

int dfs(int u, int father){ //u表示的是当前浏览到的点,father表示当前点的父节点
    int dist = 0, d1 = 0, d2 = 0;   //d1表示的是距离最大的点

    for(int i = h[u]; ~i; i = ne[i]){
        int j = e[i];
        if(j == father) continue;

        int d = dfs(j, u) + w[i];
        if(d >= d1) d2  = d1, d1 = d;  //更新最大值和第二大值
        else if(d > d2) d2 = d;     //只更新第二大值
        dist = max(dist, d);
    }
    ans = max(ans, d1 + d2);
    return dist;
}

int main(){
    cin >> n;

    memset(h, -1, sizeof h);    //初始化一定不能忘了
    for(int i = 0; i < n - 1; i ++){
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }

    dfs(1, -1);     //-1表示的是根节点的父节点


    cout << ans << endl;
    return 0;
}


活动打卡代码 AcWing 321. 棋盘分割

天命
2个月前
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 15, M = 9;
const double INF = 1e9;

int n, m = 8;
int s[M][M];
double f[M][M][M][M][N];
double X;

int get_sum(int x1, int y1, int x2, int y2)
{
    return s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1];
}

double get(int x1, int y1, int x2, int y2)
{
    double sum = get_sum(x1, y1, x2, y2) - X;
    return (double)sum * sum / n;
}

double dp(int x1, int y1, int x2, int y2, int k)
{
    double &v = f[x1][y1][x2][y2][k];
    if (v >= 0) return v;
    if (k == 1) return v = get(x1, y1, x2, y2);

    v = INF;
    for (int i = x1; i < x2; i ++ )
    {
        v = min(v, get(x1, y1, i, y2) + dp(i + 1, y1, x2, y2, k - 1));
        v = min(v, get(i + 1, y1, x2, y2) + dp(x1, y1, i, y2, k - 1));
    }

    for (int i = y1; i < y2; i ++ )
    {
        v = min(v, get(x1, y1, x2, i) + dp(x1, i + 1, x2, y2, k - 1));
        v = min(v, get(x1, i + 1, x2, y2) + dp(x1, y1, x2, i, k - 1));
    }

    return v;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= m; i ++ )
        for (int j = 1; j <= m; j ++ )
        {
            cin >> s[i][j];
            s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
        }

    X = (double)s[m][m] / n;
    memset(f, -1, sizeof f);
    printf("%.3lf\n", sqrt(dp(1, 1, 8, 8, n)));

    return 0;
}

作者:yxc
链接:https://www.acwing.com/activity/content/code/content/124467/
来源:AcWing
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。


活动打卡代码 AcWing 479. 加分二叉树

天命
2个月前
#include<iostream>

using namespace std;
const int N = 35;

int f[N][N], g[N][N];   //f表示的是区间[i, j]里面, 即中序遍历[i, j]上二叉树的集合, g表示的是区间[i, j]上根节点的编号
int n, w[N];        //记录下根节点以便以后进行前序遍历

void dfs(int l, int r)
{     //对左右区间进行枚举
    if(l > r) return ;
    int root = g[l][r];     //先找出根节点;
    cout << root << ' ' ;
    dfs(l, root - 1);
    dfs(root + 1, r);

}

int main(){
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> w[i];
    }

    for(int len = 1; len <= n; len++){
        for(int i = 1; i + len - 1 <= n; i++){
            int j = i + len - 1;
            if(len == 1){
                f[i][j] = w[i]; //如果长度为1, 这个区间内的加分就是该节点的值
                g[i][j] = i;
            }
            else {

                for(int k = i; k <= j; k ++){
                     int left = k == i? 1 : f[i][k - 1];
                     int right = k == j? 1 : f[k + 1][j];
                     int score = left * right + w[k];   //先记录下以k为分界点的子树的加分
                     if(f[i][j] < score){
                         g[i][j] = k;
                         f[i][j] = score;
                     }

                }

            }

        }
    }

    cout << f[1][n] << endl;
    dfs(1, n);

    return 0;
}



天命
2个月前
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 55, M = 35, INF = 1e9;

int n;
int w[N];
LL f[N][N][M];

void add(LL a[], LL b[])
{
    static LL c[M];
    memset(c, 0, sizeof c);
    for (int i = 0, t = 0; i < M; i ++ )
    {
        t += a[i] + b[i];
        c[i] = t % 10;
        t /= 10;
    }
    memcpy(a, c, sizeof c);
}

void mul(LL a[], LL b)
{
    static LL c[M];
    memset(c, 0, sizeof c);
    LL t = 0;
    for (int i = 0; i < M; i ++ )
    {
        t += a[i] * b;
        c[i] = t % 10;
        t /= 10;
    }
    memcpy(a, c, sizeof c);
}

int cmp(LL a[], LL b[])
{
    for (int i = M - 1; i >= 0; i -- )
        if (a[i] > b[i]) return 1;
        else if (a[i] < b[i]) return -1;
    return 0;
}

void print(LL a[])
{
    int k = M - 1;  
    while (k && !a[k]) k -- ;   //去除前导0
    while (k >= 0) cout << a[k -- ];
    cout << endl;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++ ) cin >> w[i];

    LL temp[M];
    for (int len = 3; len <= n; len ++ )
        for (int l = 1; l + len - 1 <= n; l ++ )
        {
            int r = l + len - 1;
            f[l][r][M - 1] = 1;
            for (int k = l + 1; k < r; k ++ )
            {
                memset(temp, 0, sizeof temp);
                temp[0] = w[l];
                mul(temp, w[k]);
                mul(temp, w[r]);
                add(temp, f[l][k]);
                add(temp, f[k][r]);
                if (cmp(f[l][r], temp) > 0)
                    memcpy(f[l][r], temp, sizeof temp);
            }
        }

    print(f[1][n]);

    return 0;
}


活动打卡代码 AcWing 320. 能量项链

天命
2个月前
#include<iostream>
#include<cstring>

using namespace std;

const int N = 210;
int n, m;
int f[N][N];    //f[i][j]表示的是区间[i, j]上的矩阵合并在一起的集合
int w[N];

int main(){
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> w[i];
        w[i + n] = w[i];
    }

    for(int len = 2; len <= n + 1; len ++){ //如果长度小于3, 只存在一个矩阵, 无法产生能量,不用枚举
        for(int i = 1; i + len - 1 <= n * 2; i ++){
            int j = i + len - 1;
            for(int k = i + 1; k < j; k ++){
                f[i][j] = max(f[i][j], f[i][k] + f[k][j] + w[i] * w[j] * w[k]);
            }
        }
    }

    int res = 0;
    for(int i = 1; i <= n; i ++){
        res = max(res, f[i][i + n]);
    }
    cout << res << endl;
    return 0;
}


活动打卡代码 AcWing 1068. 环形石子合并

天命
2个月前
//缺口存在的地方可能有n种,就是以1 ~ n为断点, 为了避免枚举n次,用2*n长度的线代替

#include<iostream>
#include<cstring>
using namespace std;

const int N = 410;

int f[N][N], g[N][N];   //f保存的是最大值  g保存的是最小值
int n;
int w[N];
int s[N];       //处理前缀和,可以批量对一定区间内的数进行加减

int main(){
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> w[i];
        w[i + n] = w[i];
    }
    for(int i = 1; i <= N; i++) s[i] = s[i - 1] + w[i]; 

    memset(f, -0x3f, sizeof f);
    memset(g, 0x3f, sizeof g);

    for(int len = 1; len <= n; len ++)
    {
        for(int i = 1; i + len - 1 <= 2 * n; i++)    //i表示的是左端点, j表示的是右端点
        {
            int j = i + len - 1;
            if(i == j) f[i][j] = g[i][j] = 0;
            else{
                for(int k = i; k <= j; k ++){
                    f[i][j] = max(f[i][k] + f[k + 1][j] + s[j] - s[i - 1], f[i][j]);
                    g[i][j] = min(g[i][k] + g[k + 1][j] + s[j] - s[i - 1], g[i][j]);
                }
            }

        }
    }

    int maxv = -1e9, minv = 1e9;
    for(int i = 1; i <= n; i++){
        maxv= max(maxv, f[i][i + n - 1]);
        minv = min(minv, g[i][i + n - 1]);
    }

    cout << minv << endl << maxv << endl;
    return 0;
}


活动打卡代码 AcWing 275. 传纸条

天命
2个月前
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 52;
int n, m;
int f[N * 2][N][N], w[N][N];

int main(){
    cin >> n >> m;

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            cin >> w[i][j];

    for(int i = 2; i <= m + n; i ++)
    {
        for(int x1 = max(1, i - m); x1 <= min(n, i - 1); x1 ++){
            for(int x2 = max(1, i - m); x2 <= min(n, i - 1); x2++){
                int t = w[x1][i - x1];
                if(x1 != x2) t += w[x2][i - x2];

                for(int a = 0; a <= 1; a++) 
                    for(int b = 0; b <= 1; b++){
                        f[i][x1][x2] = max(f[i][x1][x2], f[i - 1][x1 - a][x2 - b] + t);
                    }
            }       
        }
    }

    cout << f[n + m][n][n] << endl;
    return 0;
}