0924模拟赛总结

A.String Master

\(O(n^3)\)跑跑跑，至关与\(LCS\)加一维，而后考场上彷佛内存炸了，突发奇想改了\(short\)A掉了，妙啊

B.Tourist Attractions

考虑只枚举\(2-3\)这样的边，而后对答案的贡献就是，（u所链接点的个数-1） * （v所链接点的个数-1）而后再减去一个重复计算的部分，就是同时链接了\(u\)和\(v\)的点作的贡献
像这样

5显然作了两次贡献
\((d_u - 1) * (d_v - 1) - son[u]\cap son[v]\)
获得了70分的好成绩
而后用\(bitset\)维护一下取交集的操做就okk了
注意题目中较小的数据范围

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <bitset>
using namespace std;

inline int read(){
	int x = 0, w = 1;
	char ch = getchar();
	for(; ch > '9' || ch < '0'; ch = getchar()) if(ch == '-') w = -1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return x * w;
}

const int ss = 2333;

bitset<1501> b[ss];

int d[ss];
long long dp[ss][ss], ans;
char s[ss][ss];

signed main(){
	freopen("tourist.in", "r", stdin);
	freopen("tourist.out", "w", stdout);
	int n = read();
	for(register int i = 1; i <= n; i++)
		scanf("%s", s[i] + 1);
	for(register int i = 1; i <= n; i++) dp[1][i] = 1;
	for(register int i = 1; i <= n; i++)
		for(register int j = 1; j <= n; j++)
			if(s[i][j] == '1') d[i]++;
	for(register int t = 2; t <= 4; t++)
		for(register int i = 1; i <= n; i++)
			for(register int j = 1; j <= n; j++)
				if(s[i][j] == '1') dp[t][i] += dp[t - 1][j];
	for(register int i = 1; i <= n; i++)
		ans += dp[4][i];
	for(register int i = 1; i <= n; i++)
		ans -= d[i] * d[i] * 2;
	for(register int i = 1; i <= n; i++){
		for(register int j = 1; j <= n; j++){
			if(s[i][j] == '1') b[i][j] = 1, ans++;
		}
	}
	for(register int i = 1; i <= n; i++){
		for(register int j = i + 1; j <= n; j++){
			if(b[i][j]) ans -= (b[i] & b[j]).count() * 2;
		}
	}
	cout << ans << endl;
}

C.Walk

对于每个值，只有可能和本身的子集连边，而后不得不复习一下如何求一个数的全部子集

signed main(){
	int n = read();
	for(register int i = n; i; i = (i - 1) & n)
		cout << i << endl;
}