Codeforces Round #264 (Div. 2)_html/css_WEB-ITnose

Codeforces Round #264 (Div. 2)

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A: Pay attention to the situation where the special judgment is just right, and for the rest, just judge the maximum value of the record one by one

B : Scan it once, fill it with money if there is not enough, and record the excess energy

C: Process the main and secondary diagonals, and then only choose the largest one for black and white squares.

D: Transformed into a DAG longest path problem, record the position of each number in each sequence. If a number can be placed, then the number in each sequence must be at the end of the previous sequence. Behind the number

E: Work hard, pre-process the tree, and then find a matching one from the current position upwards for each query. If there is no matching one, it will be -1

Code:

A:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, s;int main() {	scanf("%d%d", &n, &s);	int x, y;	int flag = 1;	int ans = 0;	for (int i = 0; i  <br> B: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>const int N = 100005;typedef long long ll;int n;ll h[N];int main() {	scanf("%d", &n);	for (int i = 1; i  h[i - 1]) {			ll need = h[i] - h[i - 1];			if (now >= need) {				now -= need;			} else {				ans += need - now;				now = 0;			}		} else {			now += h[i - 1] - h[i];		}	}	printf("%lld\n", ans);	return 0;}</cstring></cstdio>

C:

#include <cstdio>#include <cstring>const int N = 2005;typedef long long ll;int n;ll g[N][N], zhu[N + N], fu[N + N];int main() {	scanf("%d", &n);	for (int i = 0; i  <br> D: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1005;struct Num {	int v[10], la;} num[N];bool cmp(Num a, Num b) {	return a.la  <br> E: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <vector>using namespace std;const int N = 100005;int n, q, val[N], f[N];vector<int> g[N];void dfs(int u, int fa) {	f[u] = fa;	for (int i = 0; i  1) return u;		u = f[u];	}	return -1;}int main() {	scanf("%d%d", &n, &q);	for (int i = 1; i  <br> <br> <p></p> </int></vector></cstring></cstdio>