BestCoder Round #70

最新推荐文章于 2021-08-05 19:25:09 发布

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最新推荐文章于 2021-08-05 19:25:09 发布

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原文链接：http://www.cnblogs.com/Running-Time/p/5181007.html

本文详细解析了几个C++代码片段，包括数学问题解决、背包问题优化、迷宫路径计数等算法应用，涉及从完全平方数判断到动态规划、从01背包到动态规划优化、以及迷宫路径计数的复杂问题解决。

模拟 1001 Jam's math problem

判断b ^ 2 - 4ac是否为完全平方数.当delta < 0, sqrt (delta) 输出为nan, 但是好像也能计算?

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <queue>
#include <map>
#include <cmath>

typedef long long ll;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;

int main(void)	{
	int T;	std::cin >> T;
	while (T--)	{
		ll a, b, c;	std::cin >> a >> b >> c;
		ll d = b * b - 4 * 1ll * a * c;
		if (d < 0)	puts ("NO");
		else	{
			ll e = sqrt (d);
			if (e * e == d)	puts ("YES");
			else	puts ("NO");
		}
	}

	return 0;
}
/*
long long	long double
sqrt () HDU %I64d
*/

01DP 1002 Jam's balance

原来是背包,暴力枚举是不可做的.dp[i][j] 表示前i个砝码,能称多少的重量,当然砝码也能放在另一边也就是j - w,如果j - w < 0, -(j - w)表示将物品放在堆满砝码的一边,w放在另一边.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <cmath>

int dp[22][2002];

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		int n;	scanf ("%d", &n);
		memset (dp, 0, sizeof (dp));
		dp[0][0] = 1;
		for (int w, i=1; i<=n; ++i)	{
			scanf ("%d", &w);
			for (int j=0; j<=2000; ++j)	{
				if (!dp[i-1][j])	continue;
				dp[i][j] = dp[i][j+w] = 1;
				dp[i][abs (j-w)] = 1;
			}
		}
		int m;	scanf ("%d", &m);
		for (int w, i=1; i<=m; ++i)	{
			scanf ("%d", &w);
			if (dp[n][w])	puts ("YES");
			else	puts ("NO");
		}
	}
	return 0;
}
//01dp "twice"

dp(优化) 1003 Jam's maze

这种问方案数的以后应该要想到可能是DP.f[x1][y1][x1][y1]=f[x1][y1-1][x2][y2+1]+f[x1][y1-1][x2+1][y2]+f[x1-1][y1][x2][y2+1]+f[x1-1][y1][x2+1][y2].对它优化,f[i][x1][x2],通过走的步数以及x能计算出当前走到的y,再优化点把第一维改成滚动的,即dp[now][x1][x2] <- dp[now^1][x1][x2] ... dp[now^1][x1][x2]是(x1, y1 - 1) 和 (x2, y2 - 1), 其他类似

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <cmath>

typedef long long ll;
const int N = 5e2 + 5;
const int MOD = 5201314;
int dp[2][N][N];
char str[N][N];

void add(int &a, int b)	{
	a += b;
	if (a >= MOD)	a %= MOD;
}

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		int n;	scanf ("%d", &n);
		for (int i=1; i<=n; ++i)	scanf ("%s", str[i] + 1);
		if (str[1][1] != str[n][n])	puts ("0");
		else	{
			memset (dp, 0, sizeof (dp));
			int now = 0;
			dp[now][1][n] = 1;
			for (int s=1; s<n; ++s)	{
				now ^= 1;	memset (dp[now], 0, sizeof (dp[now]));
				for (int x1=1; x1<=n; ++x1)	{
					for (int x2=n; x2>=1; --x2)	{
						if (x1 - 1 > s || n - x2 > s)	continue;
						int y1 = 1 + s - (x1 - 1);
						int y2 = n - (s - (n - x2));
						if (str[x1][y1] != str[x2][y2])	continue;
						add (dp[now][x1][x2], dp[now^1][x1][x2]);
						add (dp[now][x1][x2], dp[now^1][x1][x2+1]);
						add (dp[now][x1][x2], dp[now^1][x1-1][x2]);
						add (dp[now][x1][x2], dp[now^1][x1-1][x2+1]);
					}
				}
			}
			int ans = 0;
			for (int i=1; i<=n; ++i)	add (ans, dp[now][i][i]);
			printf ("%d\n", ans);
		}
	}

	return 0;
}

转载于:https://www.cnblogs.com/Running-Time/p/5181007.html