A - Kefa and First Steps

• 簡單dp

/* **********************************************

    File Name: a.cpp

    Auther: [email protected]

    Created Time: 2015/9/23 星期三 上午 12:31:29

*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long
 
 ll;
typedef pair<int, int> P;

const int MAX = 100007;
int a[MAX];
int dp[MAX];

int main() {
    int n;
    while (~scanf(" %d", &n)) {
        memset(dp, 0, sizeof(dp));
        a[0] = 0;
        for (int i = 1; i <= n; ++i) {
            scanf(" %d", a + i);
        }
        int ans = 1;
        for
 
 (int i = 1; i <= n; ++i) {
            if (a[i] >= a[i - 1]) {
                dp[i] = dp[i - 1] + 1;
                if (dp[i] > ans) {
                    ans = dp[i];
                }
            } else {
                dp[i] = 1;
            }
        }
        printf("%d\n", ans);
    }
    return
 
 0;
}

B - Kefa and Company

• 可以O(n∗logn)地做。當然，本題可以做到O(n)。

/* **********************************************

    File Name: b.cpp

    Auther: [email protected]

    Created Time: 2015/9/23 星期三 上午 12:37:27

*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
#ifdef ONLINE_JUDGE
#define lld I64d
#endif

const int MAX = 100007;
struct Node {
    int m;
    int s;
    bool operator<(const Node& b)const {
        return m < b.m;
    }
} a[MAX];
ll sum[MAX];

bool operator<(const ll& b, const Node& a) {
    return b < a.m;
}
bool operator<(const Node& a, const ll& b) {
    return a.m < b;
}

int main() {
    int n, d;
    while (~scanf(" %d %d", &n, &d)) {
        for (int i = 0; i < n; ++i) {
            scanf(" %d %d", &a[i].m, &a[i].s);
        }
        sort(a, a + n);
        sum[0] = a[0].s;
        for (int i = 1; i < n; ++i) {
            sum[i] = sum[i - 1] + a[i].s;
        }

        ll ans = 0LL;
        for (int i = 0; i < n; ++i) {
            ans = max(ans, sum[lower_bound(a, a + n, a[i].m + d) - a - 1] - sum[i] + a[i].s);
        }
        cout << ans << endl;
        //printf("%lld\n", ans);
    }
    return 0;
}

C - Kefa and Park

• 簡單dfs一遍即可

/* **********************************************

    File Name: c.cpp

    Auther: [email protected]

    Created Time: 2015/9/23 星期三 上午 1:07:17

*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 100007;
vector<int> G[MAX];
int cat[MAX];
bool vis[MAX];
bool is_leaf[MAX];
bool ok[MAX];
int n, m;

void dfs(int u, int now, int his) {
    vis[u] = true;

    if (cat[u] == 1) {
        ++now;
    } else {
        now = 0;
    }
    if (now > his) his = now;

    ok[u] = his <= m;
    //printf("ok[%d] = %s\n", u, ok[u] ? "true" : "false");

    int son = 0;
    for (auto it: G[u]) {
        if (vis[it]) continue;
        dfs(it, now, his);
        son++;
    }
    if (son == 0) {
        is_leaf[u] = true;
    }
}

int main() {
    while (~scanf(" %d %d", &n, &m)) {
        for (int i = 1; i <= n; ++i) {
            scanf(" %d", cat + i);
            G[i].clear();
        }
        int u, v;
        for (int i = 1; i < n; ++i) {
            scanf(" %d %d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        memset(vis, false, sizeof(vis));
        memset(ok, false, sizeof(ok));
        memset(is_leaf, false, sizeof(is_leaf));
        dfs(1, 0, 0);
        int sum = 0;
        for (int i = 1; i <= n; ++i) {
            if (is_leaf[i] && ok[i]) {
                ++sum;
            }
        }
        printf("%d\n", sum);
    }
    return 0;
}