牛客小白月赛30
A. 黑白边
题解
并查集模板题，构成最小生成树，优先选择黑边。
注意不能构成树的情况，吃了一发罚时。
代码
#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 2e5+55;
struct node {
int x, y, z;
}num[maxn];
int bin[maxn];
void init(int n) {
for (int i = 1; i <= n; ++ i) {
bin[i] = i;
}
}
int find(int x) {
return bin[x] = (x == bin[x] ? x : find(bin[x]));
}
int main() {
int n, m;
cin >> n >> m;
init(n);
for (int i = 0; i < m; ++ i) {
cin >> num[i].x >> num[i].y >> num[i].z;
}
sort(num, num+m, [](node x, node y) {
return x.z < y.z;
});
int ans = 0, cnt = n-1;
for (int i = 0; i < m; ++ i) {
int xx = find(num[i].x);
int yy = find(num[i].y);
if (xx != yy) {
if (num[i].z) ans ++;
cnt --;
bin[xx] = yy;
}
}
if (cnt) cout << -1 << endl;
else cout << ans << endl;
return 0;
} B. 最好的宝石
题解
线段树，每个节点中存放这个节点所代表的最左区间 l ，最右区间 r ， 区间 [l,r] 之间的最大值以及最大值出现的次数。
struct node {
int l, r, cnt, maxx;
}; 注意区间合并就好了吧
tree[step].maxx = max(tree[step<<1].maxx, tree[step<<1|1].maxx); if (tree[step].maxx == tree[step<<1].maxx) tree[step].cnt += tree[step<<1].cnt; if (tree[step].maxx == tree[step<<1|1].maxx) tree[step].cnt += tree[step<<1|1].cnt;
更新区间的时候需要注意的是
tree[step].cnt = 0;
代码
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e5+55;
struct node {
int l, r, cnt, maxx;
}tree[maxn<<2];
int a[maxn];
void Build(int l, int r, int step) {
tree[step].l = l;
tree[step].r = r;
tree[step].cnt = 0;
tree[step].maxx = 0;
if (l == r) {
tree[step].cnt = 1;
tree[step].maxx = a[l];
return;
}
int mid = (l+r) >> 1;
Build(l, mid, step<<1);
Build(mid+1, r, step<<1|1);
tree[step].maxx = max(tree[step<<1].maxx, tree[step<<1|1].maxx);
if (tree[step].maxx == tree[step<<1].maxx) tree[step].cnt += tree[step<<1].cnt;
if (tree[step].maxx == tree[step<<1|1].maxx) tree[step].cnt += tree[step<<1|1].cnt;
}
void Update(int pos, int val, int step) {
if (tree[step].l > pos || tree[step].r < pos) return;
if (tree[step].l == tree[step].r) {
tree[step].cnt = 1;
tree[step].maxx = val;
return;
}
Update(pos, val, step<<1);
Update(pos, val, step<<1|1);
tree[step].maxx = max(tree[step<<1].maxx, tree[step<<1|1].maxx);
tree[step].cnt = 0;
if (tree[step].maxx == tree[step<<1].maxx) tree[step].cnt += tree[step<<1].cnt;
if (tree[step].maxx == tree[step<<1|1].maxx) tree[step].cnt += tree[step<<1|1].cnt;
}
int maxx = 0, cnt = 0;
void Query(int l, int r, int step) {
if (tree[step].l > r || tree[step].r < l) return;
if (tree[step].l >= l && tree[step].r <= r) {
if (maxx == tree[step].maxx) {
cnt += tree[step].cnt;
}
if (maxx < tree[step].maxx) {
maxx = tree[step].maxx;
cnt = tree[step].cnt;
}
return;
}
Query(l, r, step<<1);
Query(l, r, step<<1|1);
}
string s;
int main() {
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; ++ i) {
cin >> a[i];
}
Build(1, n, 1);
while (m --) {
int x, y;
cin >> s >> x >> y;
if (s[0] == 'A') {
maxx = 0, cnt = 0;
Query(x, y, 1);
cout << maxx << " " << cnt << endl;
}
else {
Update(x, y, 1);
}
}
return 0;
} C. 滑板上楼梯
题解
根据贪心思路，优先走 3 台阶，那么最优的话就会形成
31313131.... 等，这样走2 步上 4 台阶，后面的进行特判即可，
代码
#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
int main() {
ll n;
cin >> n;
ll x = n/4*2;
n %= 4;
if (n == 3) x ++;
else x += n;
cout << x << endl;
return 0;
} D. GCD
题解
考虑一下最复杂的情况，根据抽屉原则，把这个区间之内的所有的质数全部放到子集 T 中，再添加一个非 1 并且没有出现过的数字，必定存在两个数的 gcd 不为 1 。
所以素数筛这个区间的素数，最后的结果加 2 即可。
注意 -1 的情况。
代码
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int PSZ = 1e7+7;
int isprimes[PSZ];
int primes[PSZ/10];
int Prime(int n) {
int cnt = 0;
isprimes[0] = isprimes[1] = 0;
for (int i = 2; i <= n; ++ i) {
isprimes[i] = 1;
}
for (int i = 2; i <= n; ++ i) {
if (isprimes[i]) {
primes[++cnt] = i;
}
for (int j = 1; j <= cnt && 1ll*primes[j]*i <= n; ++ j) {
isprimes[i*primes[j]] = 0;
if (i % primes[j] == 0) break;
}
}
for (int i = 1; i <= n; ++ i) {
isprimes[i] += isprimes[i-1];
}
return cnt;
}
int main() {
int n;
cin >> n;
int x = Prime(n);
x += 2;
if (x > n) {
cout << -1 << endl;
}
else cout << x << endl;
return 0;
} E. 牛牛的加法
题解
模拟吧，注意坑点，
- 删除前导
0 - 如果所有的结果都为
0的话，记得输出一个0
代码
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 2e5+555;
string s, t;
int a[maxn] = {0};
int main() {
cin >> s >> t;
int n = max(s.length(), t.length());
for (int i = 0; i < s.length(); ++ i) {
a[i] += s[s.length()-1-i] - '0';
}
for (int i = 0; i < t.length(); ++ i) {
a[i] += t[t.length()-1-i] - '0';
}
for (int i = 0; i < n; ++ i) {
a[i] %= 10;
}
bool flag = false;
for (int i = n-1; i >= 0; -- i) {
if (a[i]) flag = true;
if (flag) cout << a[i];
}
if (!flag) cout << 0;
cout << endl;
return 0;
} F. 石子合并
题解
首先考虑一下，最后的结果为 2n-2 个石子的和并且每个石子至少没选择的了一次。
根据这个原则的话，优先选择，除了最大的石子以外的石子都被选择一次，其余的 n-1 个石子都为最大的才是最优的结果。
代码
#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
int main() {
int n;
cin >> n;
ll maxx = 0, ans = 0, x;
for (int i = 0; i < n; ++ i) {
cin >> x;
maxx = max(maxx, x);
ans += x;
}
ans -= maxx;
ans += maxx*(n-1);
cout << ans << endl;
return 0;
} G. 滑板比赛
题解
双指针算法，模拟+贪心
对于牛妹的每个值优先选择大于它的第一个值就好了。
代码
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 2e5+55;
int a[maxn], b[maxn];
int main() {
int n, m;
cin >> n >> m;
for (int i = 0; i < n; ++ i) {
cin >> a[i];
}
for (int i = 0; i < m; ++ i) {
cin >> b[i];
}
sort(a, a+n);
sort(b, b+m);
int pos = 0;
for (int i = 0; i < n; ++ i) {
if (pos < m && a[i] > b[pos]) {
pos ++;
}
}
cout << pos << endl;
return 0;
} H. 第 k 小
题解
优先队列+模拟？
代码
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
priority_queue<int> que;
const int maxn = 2e5+55;
int a[maxn];
int main() {
int n, m, k;
cin >> n >> m >> k;
for (int i = 0; i < n; ++ i) {
cin >> a[i];
}
sort(a, a+n);
for (int i = 0; i < n && i < k; ++ i) {
que.push(a[i]);
}
int cnt = min(n, k);
while (m --) {
int x, y;
cin >> x;
if (x == 1) {
cin >> y;
cnt ++;
que.push(y);
if (cnt > k) {
cnt --;
que.pop();
}
}
else {
if (cnt == k) {
cout << que.top() << endl;
}
else cout << -1 << endl;
}
}
return 0;
} I. 区间异或
题解
枚举区间 [l,r] 之间的所有异或值。
for (int i = 0; i < n; ++ i) {
int ans = 0;
for (int j = i; j < n; ++ j) {
ans ^= a[j];
dp[i][j] = ans;
}
}
for (int i = 0; i < n; ++ i) {
for (int j = i; j < n; ++ j) {
v.push_back({dp[i][j], j-i+1});
}
} 排序+单调栈处理一下，目的是为了求出，区间 [l,r] 的最大值是多少。
sort(v.begin(), v.end(), [](node x, node y) {
return x.val < y.val;
});
for (int i = 0; i < (int)v.size(); ++ i) {
while (u.size() && v[i].site <= u[u.size()-1].site) {
u.pop_back();
}
u.push_back(v[i]);
}
for (int i = 0; i < (int)u.size(); ++ i) {
xt[i] = u[i].val;
} m 次查询，二分查找一下。
while (m --) {
int x;
cin >> x;
if (x > u[u.size()-1].val) {
cout << -1 << endl;
}
else {
int pos = lower_bound(xt, xt+u.size(), x) - xt;
cout << u[pos].site << endl;
}
} 代码
#include <queue>
#include <cmath>
#include <vector>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 3000+55;
int dp[maxn][maxn];
int a[maxn];
struct node {
int val, site;
};
vector<node> v, u;
int xt[maxn];
int main() {
int n, m;
cin >> n >> m;
for (int i = 0; i < n; ++ i) {
cin >> a[i];
}
for (int i = 0; i < n; ++ i) {
int ans = 0;
for (int j = i; j < n; ++ j) {
ans ^= a[j];
dp[i][j] = ans;
}
}
for (int i = 0; i < n; ++ i) {
for (int j = i; j < n; ++ j) {
v.push_back({dp[i][j], j-i+1});
}
}
sort(v.begin(), v.end(), [](node x, node y) {
return x.val < y.val;
});
for (int i = 0; i < (int)v.size(); ++ i) {
while (u.size() && v[i].site <= u[u.size()-1].site) {
u.pop_back();
}
u.push_back(v[i]);
}
for (int i = 0; i < (int)u.size(); ++ i) {
xt[i] = u[i].val;
}
while (m --) {
int x;
cin >> x;
if (x > u[u.size()-1].val) {
cout << -1 << endl;
}
else {
int pos = lower_bound(xt, xt+u.size(), x) - xt;
cout << u[pos].site << endl;
}
}
return 0;
} J. 小游戏
题解
dp 瞎搞。。。。。
代码
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl "\n"
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 2e5+555;
ll dp[maxn] = {0};
int main() {
int n, x;
cin >> n;
for (int i = 0; i < n; ++ i) {
cin >> x;
dp[x] += x;
}
ll maxx = 0;
for (int i = 0; i <= 200000; ++ i) {
if (i == 2) dp[i] += dp[i-2];
if (i >= 3) dp[i] += max(dp[i-2], dp[i-3]);
}
for (int i = 0; i <= 200050; ++ i) {
// cout << dp[i] << " ";
maxx = max(maxx, dp[i]);
}
// cout << endl;
cout << maxx << endl;
return 0;
} 
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