codeforces round #427 div2

Posted 2020-09-30 123456

A:读懂题，乘一下判断大小就行了

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int s, v1, v2, t1, t2;
    scanf("%d%d%d%d%d", &s, &v1, &v2, &t1, &t2);
    int ans1 = v1 * s + t1 * 2, ans2 = v2 * s + t2 * 2;
    if(ans1 == ans2) puts("Friendship");
    else if(ans1 < ans2) puts("First");
    else puts("Second");
    return 0;
}

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B:又是桶。。。cf怎么这么喜欢桶，当然把最小的位换成9是最好的，那么10个数字开10个桶，暴力删除每位改成9就行了，只是while要放在前面枚举pos

#include<bits/stdc++.h>
using namespace std;
int tot[10], sum, ans;
char s[1000010];
int main()
{
    int k;
    scanf("%d%s", &k, s);
    int len = strlen(s);
    for(int i = 0; i < len; ++i) ++tot[s[i] - \'0\'], sum += s[i] - \'0\';
    int pos = 0;
    while(sum < k)
    {
        while(tot[pos] == 0) ++pos;
        sum += 9 - pos;
        --tot[pos];
        ++ans;
    }
    printf("%d\\n", ans);
    return 0;
} 

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C:很明显不能暴力每次把所有星星+1，这是做不到的，但是看见c很小，那么我们预处理出c+1种情况就行了，可以用二维前缀和或二维bit，注意星星的位置可以重叠，考试的时候脑子坏了重新写了一遍，skip罚时爆炸，30min没了。。。

二维前缀和

#include<bits/stdc++.h>
using namespace std;
int a[11][110][110], mark[110][110];
vector<int> b[11][110][110]; 
int main()
{
    int n, q, c;
    scanf("%d%d%d", &n, &q, &c);
    for(int i = 1; i <= n; ++i) 
    {
        int x, y, s;
        scanf("%d%d%d", &x, &y, &s);
        mark[x][y] = 1;
        b[0][x][y].push_back(s);
    }
    for(int k = 0; k <= 10; ++k)
        for(int i = 1; i <= 101; ++i)
            for(int j = 1; j <= 101; ++j)
            {
                int delta = 0;
                if(k > 0)
                {
                    for(int l = 0; l < b[k - 1][i][j].size(); ++l) 
                    {        
                        b[k][i][j].push_back((b[k - 1][i][j][l] + 1) % (c + 1));
                        delta += b[k][i][j][l];
                    }
                }
                else
                {
                    for(int l = 0; l < b[k][i][j].size(); ++l)
                    delta += b[k][i][j][l]; 
                }
                a[k][i][j] = a[k][i][j - 1] + a[k][i - 1][j] - a[k][i - 1][j - 1] + delta;
            }
    for(int i = 1; i <= q; ++i)
    {
        int t, x1, y1, x2, y2;
        scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
        int sum    = a[t % (c + 1)][x2][y2] - a[t % (c + 1)][x2][y1 - 1] - a[t % (c + 1)][x1 - 1][y2] + a[t % (c + 1)][x1 - 1][y1 - 1];    
        printf("%d\\n", sum);
    }
    return 0;
}

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二维bit

#include<bits/stdc++.h>
using namespace std;
int a[11][110][110], mark[110][110];
int tree[11][110][110], x[1000010], y[1000010], s[100010][11]; 
int lowbit(int i)
{
    return i & (-i);
}
void update(int k, int x, int y, int delta)
{
    for(int i = x; i <= 101; i += lowbit(i))
        for(int j = y; j <= 101; j += lowbit(j))
            tree[k][i][j] += delta;
}
int query(int k, int x, int y)
{
    int ret = 0;
    for(int i = x; i; i -= lowbit(i))
        for(int j = y; j; j -= lowbit(j)) ret += tree[k][i][j];
    return ret;    
}
int main()
{
    int n, q, c;
    scanf("%d%d%d", &n, &q, &c);
    for(int i = 1; i <= n; ++i) 
        scanf("%d%d%d", &x[i], &y[i], &s[i][0]);
    for(int k = 0; k <= 10; ++k)
        for(int l = 1; l <= n; ++l) 
        {
            if(k == 0) update(k, x[l], y[l], s[l][0]);
            else
            {
                s[l][k] = (s[l][k - 1] + 1) % (c + 1);
                update(k, x[l], y[l], s[l][k]);
            }
        }    
    for(int i = 1; i <= q; ++i)
    {
        int t, x1, y1, x2, y2;
        scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
        int sum    = query(t % (c + 1), x2, y2) - query(t % (c + 1), x2, y1 - 1) - query(t % (c + 1), x1 - 1, y2) + query(t % (c + 1), x1 - 1, y1 - 1);    
        printf("%d\\n", sum);
    }
    return 0;
}

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bit常数真是小，跑的跟前缀和一样快

D:怎么出原题，和7D有什么区别，直接拉板子。。。dp[i]表示第i位结束的串是多少palindrome，如果是回文,dp[i]=dp[i/2]+1，否则是0，判断回文用前后哈希，就是维护一个串正反的哈希，每次O（1）维护，然后枚举起点跑dp就行了。。。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5010;
char s[N];
int dp[N], ans[N];
int main()
{
    scanf("%s", s + 1);
    int n = strlen(s + 1); 
    for(int i = 1; i <= n; ++i)
    {
        memset(dp, 0, sizeof(dp));
        ll p = 1, pre = 0, bak = 0;
        for(int j = i; j <= n; ++j)
        {
            pre = pre * 1234567ll + s[j];
            bak = s[j] * p + bak;
            p *= 1234567ll;
            if(pre == bak) dp[j - i + 1] = dp[(j - i + 1) / 2] + 1;
            ++ans[dp[j - i + 1]];
        }
    }
    for(int i = n; i; --i) ans[i] += ans[i + 1];
    for(int i = 1; i <= n; ++i) printf("%d ", ans[i]);
    return 0;
}

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F:比较套路没想出来。。。zz。。。看见基环树，我们就要把环展开，具体方式是cycle存了环上每个点，最后push_back(cycle[0]),这样就首尾相接了，然后对于环上每个点维护max_d,挂在这个点下的最长链，pre_dia,从1->i这段环，包括环上点挂着的链的直径，pre_len,1->i这段环到1的最长路径，bak也一样

 

具体是这个样子的，然后枚举分割点，把环分割成两段后，mn=min(mn,max(pre_len+bak_len,max(pre_dia,bak_dia))，两段len相加构成了一条完全的链，两个dia分别是这两段的直径。

ans=max(ans,mn) ans先预处理不在环上的最长距离，就是子树内的直径，环上分割不影响这一段，所以和mn取max。注意最后push的cycle[0] max_d要取0，否则可能cycle在第一段和第二段都取了cycle[0],这样就不对了，但是后面一段的dia要加上max_d，因为两段dia是分别互不影响的，这个东西坑了我们长时间，又没办法对拍。。。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200010;
inline int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < \'0\' || c > \'9\') { if(c == \'-\') f = -1; c = getchar(); }
    while(c >= \'0\' && c <= \'9\') { x = x * 10 + c - \'0\'; c = getchar(); }
    return x * f;
}
struct edge {
    int to, w;
    edge(int to = 0, int w = 0) : to(to), w(w) {}
};
ll ans = -1;
int n, dfn_clock, tar; 
vector<edge> G[N];
vector<int> cycle;
int dfn[N], mark[N]; 
ll w[N], pre_len[N], bak_len[N], pre_dia[N], bak_dia[N], max_d[N];
bool dfs(int u, int last)
{
    if(dfn[u]) 
    {
        tar = u;
        return true;
    }
    dfn[u] = ++dfn_clock;
    for(int i = 0; i < G[u].size(); ++i)
    {
        edge e = G[u][i];
        if(e.to == last) continue;
        if(dfs(e.to, u)) 
        {
            if(dfn[u] >= dfn[tar]) 
            {
                cycle.push_back(u);
                mark[u] = 1;
            }
            return true;
        }
    }
    return false;
}
ll Dfs(int u, int last)
{
    ll max_d = 0;
    for(int i = 0; i < G[u].size(); ++i)
    {
        edge e = G[u][i];
        if(e.to == last || mark[e.to]) continue;
        ll dep = Dfs(e.to, u);
        ans = max(ans, dep + max_d + (ll)e.w);
        max_d = max(max_d, dep + (ll)e.w);
    }
    return max_d;
}
int main()
{
    n = read();
    for(int i = 1; i <= n; ++i) 
    {
        int u = read(), v = read(), w = read();
        G[u].push_back(edge(v, w));
        G[v].push_back(edge(u, w));
    }
    dfs(1, 0);
    reverse(cycle.begin(), cycle.end());
    cycle.push_back(cycle[0]);
    for(int i = 0; i < cycle.size() - 1; ++i)
    {
        int u = cycle[i], v = cycle[i + 1];
        for(int j = 0; j < G[u].size(); ++j) 
        {
            edge e = G[u][j];
            if(e.to == v) 
            {
                w[i] = e.w;
                break;
            }
        }
        max_d[i] = Dfs(u, 0);
    }
    max_d[cycle.size() - 1] = max_d[0];
    ll cur_len = 0, cur_dia = max_d[0];
    pre_len[0] = pre_dia[0] = max_d[0];
    for(int i = 1; i < cycle.size(); ++i) 
    {
        cur_len += w[i - 1];
        cur_dia += w[i - 1];
        pre_len[i] = max(pre_len[i - 1], cur_len + max_d[i]);
        pre_dia[i] = max(pre_dia[i - 1], cur_dia + max_d[i]);
        cur_dia = max(cur_dia, max_d[i]);
    }
    cur_len = 0;
    cur_dia = max_d[cycle.size() - 1];
    bak_dia[cycle.size() - 1] = max_d[cycle.size() - 1];
    for(int i = cycle.size() - 2; i >= 0; --i) 
    {
        cur_len += w[i];
        cur_dia += w[i];
        bak_len[i] = max(bak_len[i + 1], cur_len + max_d[i]);
        bak_dia[i] = max(bak_dia[i + 1], cur_dia + max_d[i]);
        cur_dia = max(cur_dia, max_d[i]);
    }
    ll mn = 1000000000000000ll;
    for(int i = 0; i < cycle.size() - 1; ++i)
    {
        mn = min(mn, max(pre_len[i] + bak_len[i + 1], max(pre_dia[i], bak_dia[i + 1])));
//        ans = min(ans, pre_len[i] + bak_len[i + 1]); 
    }
    printf("%lld\\n", max(ans, mn));
    return 0;
}

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