Codeforces Round 966 (Div. 3)
A. Primary Task
if-else语句练习，判断前两个字符是否为"10"，并判断之后的字符是否大于1
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 1e4 + 5;
char s[N];
int main()
{
int T = read();
while(T--)
{
scanf("%s", s + 1);
int len = strlen(s + 1);
if(s[1] == '1' && s[2] == '0' && ((len > 3 && s[3] >= '1') || (len == 3 && s[3] >= '2'))) printf("YES\n");
else printf("NO\n");
}
return 0;
}
B. Seating in a Bus
模拟
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 2e5 + 5;
int a[N], vis[N];
int main()
{
int T = read();
while(T--)
{
int n = read();
for(int i = 1; i <= n; ++i) a[i] = read();
for(int i = 1; i <= n; ++i) vis[i] = 0;
vis[a[1]] = 1;
int flag = 1;
for(int i = 2; i <= n; ++i)
{
int ner = 0;
if(a[i] > 1) ner |= vis[a[i] - 1];
if(a[i] < n) ner |= vis[a[i] + 1];
if(ner == 0){ flag = 0; break; }
vis[a[i]] = 1;
}
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}
C. Numeric String Template
还是模拟，每个数字第一次出现映射到一个字符上，之后再出现直接判断，字符第一次出现同理
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 2e5 + 5;
int a[N], b[N];
int id1[N], id2[N];
char s[N];
void solve()
{
int n = read();
for(int i = 1; i <= n; ++i) a[i] = b[i] = read();
sort(b + 1, b + n + 1);
int x = unique(b + 1, b + n + 1) - (b + 1);
for(int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + x + 1, a[i]) - b;
int m = read();
while(m--)
{
scanf("%s", s + 1);
int flag = 1, len = strlen(s + 1);
if(len != n){ printf("NO\n"); continue; }
for(int i = 1; i <= n; ++i) id1[i] = -1;
for(int i = 1; i <= n; ++i)
{
if(id1[a[i]] == -1) id1[a[i]] = s[i] - 'a';
else if(id1[a[i]] != s[i] - 'a') flag = 0;
if(!flag) break;
}
for(int i = 0; i <= 25; ++i) id2[i] = -1;
for(int i = 1; i <= len; ++i)
{
if(id2[s[i] - 'a'] == -1) id2[s[i] - 'a'] = a[i];
else if(id2[s[i] - 'a'] != a[i]) flag = 0;
if(!flag) break;
}
if(flag) printf("YES\n");
else printf("NO\n");
}
}
int main()
{
int T = read();
while(T--) solve();
return 0;
}
D. Right Left Wrong
注意到 \(LLRR\) 取 \((1, 3)(2, 4)\) 和 \((1, 4)(2, 3)\) 是等价的，于是直接取所有区间只有包含没有相交的情况，即用双指针从两端向中间扫即可
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 2e5 + 5;
int a[N];
char s[N];
ll sum[N];
void solve()
{
int n = read();
for(int i = 1; i <= n; ++i) a[i] = read(), sum[i] = sum[i - 1] + a[i];
scanf("%s", s + 1);
int l = 1, r = n;
ll ans = 0;
while(l < r)
{
while(l <= n && s[l] != 'L') ++l;
while(r >= 1 && s[r] != 'R') --r;
if(l < r) ans += sum[r] - sum[l - 1];
++l, --r;
}
printf("%lld\n", ans);
}
int main()
{
int T = read();
while(T--) solve();
return 0;
}
E. Photoshoot for Gorillas
简单计算一下一个格子被多少个边长为 \(K\) 的子方阵包含，那么这个格子里的数就会贡献这么多次，贪心的让大数放到次数多的格子里
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 2e5 + 5;
int a[N];
int cnt[N];
bool cmp(int a, int b){ return a > b; }
void solve()
{
int n = read(), m = read(), K = read(), W = read(), tot = 0;
for(int i = 1; i <= W; ++i) a[i] = read();
sort(a + 1, a + W + 1, cmp);
for(int x = 1; x <= n; ++x)
for(int y = 1; y <= m; ++y)
{
int lx = max(1, x - K + 1), rx = min(n - K + 1, x);
int ly = max(1, y - K + 1), ry = min(m - K + 1, y);
cnt[++tot] = 1ll * (rx - lx + 1) * (ry - ly + 1);
}
sort(cnt + 1, cnt + tot + 1, cmp);
ll ans = 0;
for(int i = 1; i <= W; ++i) ans += 1ll * a[i] * cnt[i];
printf("%lld\n", ans);
}
int main()
{
int T = read();
while(T--) solve();
return 0;
}
F. Color Rows and Columns
首先想到的是贪心，即一个 \(a \times b\) 的矩形，填满一行或一列需要 \(\min\{a, b\}\) 步
这样贪心的取，发现当 \(1 \times 1\) 时一次性填了一行和一列，答案 + 2
贪心不完全对，转而考虑背包 \(DP\)
需要注意的是：想要答案 + a + b ，需要 \(a \times b\) 步，想要答案 + a + b - 1 ，也需要 \(a \times b\) 步
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int inf = 0x3f3f3f;
int dp[2][105], a[1005], b[1005];
vector< pair<int, int> > x[1005];
void init(int id)
{
x[id].clear();
int A = a[id], B = b[id], sum = 0, K = 0;
while(A && B)
{
sum += A, --B, ++K;
if(A > B) swap(A, B);
if(A == 0) x[id].emplace_back(pair<int, int>(K, sum)), ++K;
x[id].emplace_back(pair<int, int>(K, sum));
}
}
void solve()
{
int n = read(), K = read();
for(int i = 1; i <= n; ++i)
{
a[i] = read(), b[i] = read();
if(a[i] > b[i]) swap(a[i], b[i]);
init(i);
}
for(int i = 1; i <= K; ++i) dp[0][i] = inf;
int ans = inf;
for(int i = 1, op = 1; i <= n; ++i, op ^= 1)
{
for(int j = 0; j <= K; ++j) dp[op][j] = dp[op ^ 1][j];
for(auto [w, val] : x[i])
{
for(int j = w; j <= K; ++j)
dp[op][j] = min(dp[op][j], dp[op ^ 1][j - w] + val);
}
ans = min(ans, dp[op][K]);
}
if(ans == inf) printf("-1\n");
else printf("%d\n", ans);
}
int main()
{
int T = read();
while(T--) solve();
return 0;
}
G. Call During the Journey
先二分答案，之后是最短路魔改，松弛操作时分三种情况：
-
直接走路过去，不受任何限制
-
坐车过去，需要保证下车时间 \(\le t1\) 或者上车时间 \(\ge t2\)
-
等到 \(t2\) 再上车
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const ll inf = 0x7fffffffffff;
const int N = 1e5 + 5;
int n, m;
int t0, t1, t2;
struct node
{
int v, l1, l2;
node(){ v = l1 = l2 = 0; }
node(int x, int y, int z){ v = x, l1 = y, l2 = z; }
};
vector<node> e[N];
ll dis[N];
bool vis[N];
priority_queue< pair<int, int> > q;
bool check(int t)
{
// printf("t = %d\n", t);
for(int i = 1; i <= n; ++i) dis[i] = inf, vis[i] = 0;
dis[1] = t;
q.push(pair<int, int>(-dis[1], 1));
while(!q.empty())
{
int u = q.top().second ;
q.pop();
if(vis[u]) continue;
vis[u] = 1;
// printf("u = %d, dis = %lld\n", u, dis[u]);
for(auto [v, l1, l2] : e[u])
{
int flag = 0;
if(dis[v] > dis[u] + l2)
{
dis[v] = dis[u] + l2;
flag = 1;
}
if(dis[v] > dis[u] + l1 && ((dis[u] + l1 <= t1) || (dis[u] >= t2)))
{
dis[v] = dis[u] + l1;
flag = 1;
}
if(dis[u] <= t2 && dis[v] > t2 + l1)
{
dis[v] = t2 + l1;
flag = 1;
}
if(flag) q.push(pair<int, int>(-dis[v], v));
}
}
if(dis[n] <= t0) return true;
else return false;
}
void solve()
{
n = read(), m = read();
t0 = read(), t1 = read(), t2 = read();
for(int i = 1; i <= n; ++i) e[i].clear();
for(int i = 1; i <= m; ++i)
{
int u = read(), v = read(), l1 = read(), l2 = read();
e[u].emplace_back(node(v, l1, l2));
e[v].emplace_back(node(u, l1, l2));
}
int l = -1, r = t0;
while(l < r)
{
int mid = (l + r + 1) >> 1;
if(check(mid)) l = mid;
else r = mid - 1;
}
printf("%d\n", l);
}
int main()
{
int T = read();
while(T--) solve();
return 0;
}
H. Ksyusha and the Loaded Set
转化题意为，最初所有数都为 \(1\) ，集合中包含的数为 \(0\) ，三种操作：
-
加入集合一个数，等价于将此位置设为 \(0\)
-
删去集合一个数，等价于将此位置设为 \(1\)
-
询问集合中1的连续段长度 \(\ge k\) 的最小左端点
权值线段树搞就完了
终于知道多测的意义了，专门卡 \(O(值域)\) 的算法
注意到每次测试时都重建一颗一模一样的树，时间消耗主要在这里，不优
又注意到每个测试中的操作数是 \(O(n)\) 的，考虑撤销操作
只在一开始建树，之后每次用时撤销前面的测试的操作
总结
- 权值线段树遇上多测，考虑撤销操作
点击查看代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
int read()
{
int x = 0, f = 0; char c = getchar();
while (c < '0' || c > '9') f = c == '-', c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 2e5 + 5;
int n, m, a[N];
struct node
{
int mx, lsum, rsum, len;
node(){ mx = lsum = rsum = len = 0; }
node friend operator + (node a, node b)
{
node c;
c.mx = max(a.mx , b.mx );
c.mx = max(c.mx , a.rsum + b.lsum );
c.len = a.len + b.len ;
c.lsum = a.lsum + ((a.lsum == a.len ) ? b.lsum : 0);
c.rsum = b.rsum + ((b.rsum == b.len ) ? a.rsum : 0);
return c;
}
};
#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)
node A[8388605 + 10];
void build(int k, int l, int r)
{
if(l == r)
{
A[k].mx = A[k].lsum = A[k].rsum = A[k].len = 1;
return ;
}
int mid = (l + r) >> 1;
build(ls(k), l, mid), build(rs(k), mid + 1, r);
A[k] = A[ls(k)] + A[rs(k)];
}
void update(int k, int l, int r, int pos, int val)
{
if(l == r)
{
A[k].mx = A[k].lsum = A[k].rsum = val;
return ;
}
int mid = (l + r) >> 1;
if(pos <= mid) update(ls(k), l, mid, pos, val);
else update(rs(k), mid + 1, r, pos, val);
A[k] = A[ls(k)] + A[rs(k)];
}
int query(int k, int l, int r, int len)
{
if(l == r) return l;
int mid = (l + r) >> 1;
if(A[ls(k)].mx >= len) return query(ls(k), l, mid, len);
if(A[ls(k)].rsum + A[rs(k)].lsum >= len) return mid - A[ls(k)].rsum + 1;
return query(rs(k), mid + 1, r, len);
}
pair<int, int> sta[N << 1];
int top;
void solve()
{
n = read();
for(int i = 1; i <= n; ++i) a[i] = read();
m = read();
top = 0;
for(int i = 1; i <= n; ++i) update(1, 1, 4e6, a[i], 0), sta[++top] = pair<int, int>(a[i], 0);
while(m--)
{
char c;
int x;
scanf(" %c %d", &c, &x);
if(c == '-') update(1, 1, 4e6, x, 1), sta[++top] = pair<int, int>(x, 1);
if(c == '+') update(1, 1, 4e6, x, 0), sta[++top] = pair<int, int>(x, 0);
if(c == '?') printf("%d ", query(1, 1, 4e6, x));
}
while(top)
{
update(1, 1, 4e6, sta[top].first , sta[top].second ^ 1);
--top;
}
printf("\n");
}
int main()
{
int T = read();
build(1, 1, 4e6);
while(T--) solve();
return 0;
}
