CF-595

题目传送门

A .Yet Another Dividing into Teams

sol：原先是用比较复杂的方法来解的，后来学弟看了一眼，发现不是1就是2，当出现两个人水平相差为1就分成两组，1组全是奇数，1组全是偶数，必然符合题意。

  • 思维 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 110;
bool vis[MAXN];
int main() {
    int t, n, m;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        bool ok = true;
        memset(vis, false, sizeof(vis));
        for (int i = 1; i <= n; i++) {
            scanf("%d", &m);
            vis[m] = true;
            if (vis[m - 1] || vis[m + 1]) ok = false;
        }
        if (ok) puts("1");
        else puts("2");
    }
    return 0;
}
    View Code

     

B. Books Exchange

sol：很裸的并查集，i和p[i]属于同一个集合，求每个点所在集合的大小

  • 并查集 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 2e5 + 10;
int pre[MAXN], cnt[MAXN];
int find(int i) {
    if (pre[i] == i) return i;
    return pre[i] = find(pre[i]);
}
int main() {
    int t, n, m;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            cnt[i] = 1;
            pre[i] = i;
        }
        for (int i = 1; i <= n; i++) {
            scanf("%d", &m);
            int u = find(i);
            int v = find(m);
            if (u != v) {
                pre[u] = v;
                cnt[v] += cnt[u];
            }
        }
        for (int i = 1; i <= n; i++)
            printf("%d ", cnt[find(i)]);
        puts("");
    }
    return 0;
}
    View Code

     

C1. Good Numbers (easy version)

sol：由于n比较小，所以我们可以暴力罗列good number，pow(3, 10)已经大于n的最大范围10000，所以罗列到3的10次方够了，再组合一下，排序后用二分去搜就好了。

  • 暴力组合 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
int p[20] = {1};
vector<int> v;
void init() {
    for (int i = 1; i <= 10; i++)
        p[i] = 3 * p[i - 1];
    for (int i = 1; i <= (1 << 10); i++) {
        int sum = 0;
        for (int j = 0; j <= 10; j++)
            if (i & (1 << j)) sum += p[j];
        v.push_back(sum);
    }
}
int main() {
    int t, n;
    scanf("%d", &t);
    init();
    sort(v.begin(), v.end());
    while (t--) {
        scanf("%d", &n);
        printf("%d\n", *lower_bound(v.begin(), v.end(), n));
    }
    return 0;
}
    View Code

     

C2. Good Numbers (hard version)

sol：细看这题感觉和进制有关，把n转成3进制后找到第一个大于等于n并且所以三进制位不为2的数；

  • 进制 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
int p[60];
int main() {
    int t;
    LL n, pos;
    scanf("%d", &t);
    while (t--) {
        scanf("%lld", &n);
        memset(p, 0, sizeof(p));
        pos = 0;
        while (n) {
            p[pos++] = n % 3;
            n /= 3;
        }
        int k = pos - 1;
        while (k > 0 && p[k] != 2) k--; // 从高位往低位搜找到第一个2
        for (int i = k - 1; i >= 0; i--) p[i] = 0; // 第k位加1,那么k之前的位可以全部为0了,一定不会小于n
        while (p[k] == 2) { //进位
            p[k] = 0;
            p[++k] ++;
        }
        pos = max(pos, k + 1LL);
        LL sum = 0, pp = 1;
        for (int i = 0; i < pos; i++) {
            if (p[i] == 1) sum += pp;
            pp *= 3;
        }
        printf("%lld\n", sum);
    }
    return 0;
}
    View Code

     

D. Too Many Segments 

sol：用差分确定覆盖每个点的线段有多少条，对于点i，如果覆盖的线段数量大于k，删掉覆盖该点的右端点最远的线段，可以用优先队列找这样的线段。就是需要两个排序方式。

  • 贪心，差分 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 2e5 + 10;
struct Seg {
    int l, r;
    int index;
    friend bool operator < (Seg a, Seg b) {return a.r < b.r;}
} a[MAXN];
priority_queue<Seg> que;
bool cmp(Seg a, Seg b) {return a.l < b.l;}
vector<int> ans;
int cnt[MAXN], pos, sum;
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &a[i].l, &a[i].r);
        a[i].index = i;
    }
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i++) {
        while (pos < a[i].l) sum += cnt[++pos];
        que.push(a[i]);
        sum ++, cnt[a[i].r + 1] --;
        if (sum > m) {
            sum --;
            Seg s = que.top();
            que.pop();
            cnt[s.r + 1] ++;
            ans.push_back(s.index);
        }
    }
    printf("%d\n", ans.size());
    for (auto index : ans) printf("%d ", index);
    return 0;
}
    View Code

     

E. By Elevator or Stairs?

sol：dp，dp[0][i]表示从a[i]到1的最少时间，dp[1][i]表示从b[i]到1的最少时间。代码中直接把a、b数组当dp[0]和dp[1]来用了；

  • 动态规划 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 2e5 + 10;
int a[MAXN], b[MAXN];
int main() {
    int n, c;
    scanf("%d%d", &n, &c);
    a[1] = 0, b[1] = c;
    for (int i = 2; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 2; i <= n; i++) scanf("%d", &b[i]);
    for (int i = 1; i <= n; i++) {
        a[i] += min(a[i - 1], b[i - 1]);
        b[i] += min(a[i - 1] + c, b[i - 1]);
        printf("%d ", min(a[i], b[i]));
    }
    return 0;
}
    View Code

     

F. Maximum Weight Subset

sol：树形dp，dp[u][dep]表示u子树中距离u最近的选取点离u的距离不小于dep的最大权重。要注意的是合并的时候两颗子树间的选取点距离也不能小于等于k；

  • 树形dp 
    #include "bits/stdc++.h"
using namespace std;
#define debug puts("what the fuck");
typedef long long LL;
typedef pair<int, int> PII;
const int MAXN = 210;
int w[MAXN];
int dp[MAXN][MAXN];
vector<int> edge[MAXN];
int n, k;
void dfs(int u, int fa) {
    for (auto v : edge[u]) {
        if (v == fa) continue;
        dfs(v, u);
    }
    dp[u][0] = w[u];
    for (auto v : edge[u]) {
        if (v == fa) continue;
        dp[u][0] += dp[v][k];
    }
    for (int dep = 1; dep <= k; dep++) {
        for (auto v : edge[u]) {
            if (v == fa) continue;
            int sum = dp[v][dep - 1];
            for (auto other : edge[u]) {
                if (other == fa || other == v) continue;
                sum += dp[other][max(dep - 1, k - dep)];
            }
            dp[u][dep] = max(dp[u][dep], sum);
        }
    }
    for (int i = k - 1; i >= 0; i--) dp[u][i] = max(dp[u][i], dp[u][i + 1]);
}
int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
    for (int i = 2; i <= n; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    dfs(1, -1);
    printf("%d\n", dp[1][0]);
    return 0;
}
    View Code

    官方题解里vector太多了，看的好懵逼。提交最后解释了一下为什么复杂度是O(n^3)不是O(n^4)，但是我看着还是感觉复杂度是O(n^4)

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第一次把一场CF所有题都补出来，开心

posted @ 2019-10-24 15:06  Jathon-cnblogs  阅读(555)  评论(0)  编辑  收藏  举报