洛谷P1052过河

题目

不看数据范围的话是一个很简单的DP，可是加上数据范围之后就之前的做法就不行了。

所以我们考虑一下路径压缩。

小数据Code

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int l, m, s, t, dp[100010];
int flag[100010];
int main()
{
    scanf("%d%d%d%d", &l, &s, &t, &m);
    for (int i = 1; i <= m; i++)
    {
        int a;
        scanf("%d", &a);
        flag[a] = 1;
    }
    for (int i = 1; i <= l; i++)
    {
    	dp[i] = 21474836;
        for (int j = max(i - t, 0); j <= i - s; j++)
            dp[i] = min(dp[i], dp[j] + flag[i]);
    }
    printf("%d", dp[l]);
}

大数据Code

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#define ha s * t
#define N 100111
#define inf 214748364
#define int long long
using namespace std;
int l, m, s, t, dp[N], ans = inf * 10, r;
int data[N], a[N], b[N];
inline void init()
{
	scanf("%lld%lld%lld%lld", &l, &s, &t, &m);
    for (int i = 1; i <= m; i++)
    	cin >> data[i];
    sort(data + 1, data + 1 + m);
	if (s == t)
	{
		int ans = 0;
		for (int i = 1; i <= m; i++)
			if (!(data[i] % s)) ans++;
		printf("%lld", ans);
		exit(0);
	}		
	for (int i = 1; i <= m; i++)
	{
		int cha = data[i] - data[i - 1];
		if (cha >= ha) cha = ha;
		a[i] = a[i - 1] + cha;
		b[a[i]] = 1;
	}
	r = a[m] + ha;
	for (int i = 1; i <= r; i++)
		dp[i] = inf;
}
signed main()
{
	init();
	for (int i = 1; i <= r; i++)
		for (int j = s; j <= t; j++)
		{
			if (i >= j)
			{
				if (b[i]) dp[i] = min(dp[i - j] + 1, dp[i]);//如果该数已被标记，就要加1 
				else dp[i] = min(dp[i - j], dp[i]);	
			}
		}
	for (int i = a[m]; i <= r; i++)
		ans = min(ans, dp[i]);
	printf("%lld", ans);//输出最小值。 
	return 0;
}