10.31 模拟赛

A LIS

考虑每个数字前从$m$降序构造到$a_i$即可。

#include <iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
#define MAXN 300006
int n , m , k;
int A[MAXN];
vector<int> ans;
int main() {
	cin >> n >> m >> k;
	for( int i = 1 ; i <= k ; ++ i ) scanf("%d",&A[i]);
	int lef = n - k;
	for( int i = 1 ; i <= k ; ++ i ) {
		int cur = m;
		while( lef && cur > A[i] ) ans.push_back(cur), -- cur ,-- lef; 
		ans.push_back( A[i] );
	}
	if( lef ) return puts("No") , 0;
	puts("Yes");
	for( auto it : ans ) printf("%d ",it);
}

T2 图论

看到数据范围考虑暴搜，枚举答案（其实也就$2n$）只要倒着枚举以前的边就显然仍然存在。

于是每次加边后进行一下宽搜，每个边只会入队一次，每次扩展是$O(n)$故总复杂度$O(n^3) + O(n^3)$

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define MAXN 1006
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
int n , m , cnt;
int deg[MAXN] , G[MAXN][MAXN];
queue<pii> Q;
void bfs( int x ) {
	for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            if (!G[i][j] && deg[i] + deg[j] >= x) Q.push(make_pair(i, j)), G[i][j] = G[j][i] = 1;
    while (!Q.empty()) {
        pii now = Q.front(); Q.pop();
        ++ deg[now.fi] , ++ deg[now.se];
        ++ cnt;
        int u = now.first, v = now.second;
        for (int i = 1; i <= n; i++)
            if (i != u && !G[u][i] && deg[u] + deg[i] >= x)
                G[u][i] = G[i][u] = 1 , Q.push(mp(u, i));
        for (int i = 1; i <= n; i++)
            if (i != v && !G[v][i] && deg[v] + deg[i] >= x)
                G[v][i] = G[i][v] = 1 , Q.push(mp(v, i));
    }
}
int main() {
	cin >> n >> m;
    for (int i = 1 , u , v; i <= m; ++ i) 
        scanf("%d%d", &u, &v) , deg[u]++, deg[v]++ , G[u][v] = G[v][u] = 1 , cnt++;
    int all = n * ( n - 1 ) / 2;
    for (int i = 2 * n; i >= 0; -- i) {
        bfs( i );
        if (cnt == all) 
            return printf("%d\n", i) , 0;
    }
}

T3 防御

考虑每个点通过这堵墙投影到$x$轴上一段区间，于是就成了一个区间包含问题（二维偏序）

然后就鸽掉了。。精度误差毒瘤。