NOIP做题练习（day2）

A - Reign

题面

题解

最大子段和+\(DP\)。

预处理两个数组：

• \(p[i]\)表示 \(i\) 之前的最大子段和。
• \(l[i]\)表示 \(i\) 之后的最大子段和。

最后直接输出即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int t, n, m, k, ans, a[100003], p[100003], l[100003];

signed main()
{
	//File("REIGN");
	t = gi();
	while (t--)
	{
		n = gi(), k = gi();
		for (itn i = 1; i <= n; i+=1) a[i] = gi();
		memset(p, 0, sizeof(p)); memset(l, 0, sizeof(l));
		ans = -0x7fffffff;
		for (itn i = 1; i <= n; i+=1)
		{
		    if (p[i - 1] < 0) p[i] = a[i];
			else p[i] = p[i - 1] + a[i];
		}
		p[0] = -0x7fffffff;
		for (itn i = 1; i <= n; i+=1) p[i] = max(p[i], p[i - 1]);
		for (int i = n; i >= 1; i-=1)
		{
			if (l[i + 1] < 0) l[i] = a[i];
			else l[i] = l[i + 1] + a[i];
		}
		l[n + 1] = -0x7fffffff;
		for (itn i = n; i >= 1; i-=1) l[i] = max(l[i], l[i + 1]);
		for (int i = 1; i < n - k; i+=1) ans = max(ans, p[i] + l[i + k + 1]);
		printf("%lld\n", ans);
	}
	return 0;
}

C - Strongly Connected City

题面

题意简述

给定\(n\)条水平的单向街道和\(m\)条竖直单向的街道，其交点共有\(n \times m\)个，问这些节点是否都能互相到达。

\(2 \leq n,m \leq 20\)

题解

只需要判断最外圈是不是一个环即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <string>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

bool fl = false;
int n, m, a[25][25][2], tot, vis[25][25];
char s[2][25];

int main()
{
	n = gi(), m = gi();
	string s[3];
	cin >> s[1];
	if (s[1][0] == s[1][n - 1]) {puts("NO"); return 0;}
	cin >> s[2];
	if (s[1][0] == '<' && s[2][0] == '^') {puts("NO"); return 0;}
    if (s[1][0] == '>' && s[2][m - 1] == '^') {puts("NO"); return 0;}
    if (s[1][n - 1] == '<' && s[2][0] == 'v') {puts("NO"); return 0;}
    if (s[1][n - 1] == '>' && s[2][m - 1] == 'v') {puts("NO"); return 0;}
	puts("YES");
	return 0;
}

F - Chef and Digit Jumps

题面

题解

\(BFS\)裸题。

有一个优化：让每一个点在队列中只会出现一次，优化时间复杂度。

注意一些细节。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <string>
#include <vector>
#include <queue>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

char s[100003];
int n, m, len, p, ans, sum, b[1000003], vis[100003];
vector <int> a[15];
queue <int> q;

int main()
{
	//File("DIGJUMP");
	scanf("%s", s + 1);
	len = strlen(s + 1);
	for (int i = 1; i <= len; i+=1)
	{
		a[s[i] - '0'].push_back(i);
	}
	q.push(1); b[1] = 0;
	while (!q.empty())
	{
		int x = q.front(); q.pop();
		if (x == len) break;
		int y = s[x] - '0';
		if (!vis[y])
		{
			vis[s[x] - '0'] = 1;
			for (int i = 0; i < a[y].size(); i+=1)
			{
				int z = a[y][i];
				if (z != x && b[z] == 0)
				{
					b[z] = b[x] + 1;
					q.push(z);
				}
			}
		}
		if (x - 1 >= 1 && b[x - 1] == 0) {b[x - 1] = b[x] + 1; q.push(x - 1);}
		if (x + 1 <= len && b[x + 1] == 0) {b[x + 1] = b[x] + 1; q.push(x + 1);}
	}
	printf("%d\n", b[len]);
	return 0;
}

总结

这次做题做得很不理想。

\(T3\)时间复杂度算错浪费了很多时间。

\(T1\)想了很久才想到正解。

要多多练习，才能有提高啊！