洛谷P1514引水入城

题目

搜索加贪心其实并不需要用到\(DP\),搜索也是比较简单地搜索。

对于每个第一行的城市进行类似于滑雪那道题的搜索，然后记录最后一行它所覆盖的区间，易得一个一行城市只会有一个区间。然后可以在最后进行线段覆盖贪心即可求出答案。要注意区间闭开和边界问题。

\(Code\)

#include <bits/stdc++.h>
#define N 501
using namespace std;
int di[5] = {0, 1, -1, 0, 0}; int dj[5] = {0, 0, 0, 1, -1};
int n, m, flag, tot, data[N][N], vis[N], dp[N][N];
struct C7 {
	int i, j;
};
struct block {
	int l, r;
}s[100010];
bool cmp(block a, block b)
{
	if (a.l == b.l)
		return a.r > b.r;
	return a.l < b.l;
}
inline void init()
{
 	scanf("%d%d", &n, &m);
 	for (int i = 1; i <= n; i++)
 		for (int j = 1; j <= m; j++)
  			scanf("%d", &data[i][j]);
}
inline void bfs()
{
 	queue <C7> q;
 	for (int o = 1; o <= m; o++)
 	{
 		memset(dp, 0, sizeof(dp));
 		if (n == 1)	vis[o] = 1;
 		dp[1][o] = o;
 		q.push({1, o});
 		while (!q.empty())
 		{
 		 	C7 cur = q.front(); q.pop();
 		 	int i = cur.i, j = cur.j;
		 	for (int k = 1; k <= 4; k++)
		 	{
		 	 	int nexi = i + di[k], nexj = j + dj[k];
		 	 	if (nexi <= 0 || nexj <= 0 || nexi > n 	|| nexj > m)
		 		 	continue;
		 	 	if (!dp[nexi][nexj] && data[nexi][nexj] < data[i][j])
		 	 	{
		 		 	dp[nexi][nexj] = o;
		 		 	q.push({nexi, nexj});
		 		 	if (nexi == n)
		 	 	 		vis[nexj] = 1;
		 	 	}
		 	}
	 	}
	 	flag = 0;
	 	for (int i = 1; i <= m + 1; i++)
	 	{
		 	if (dp[n][i] && !flag)
		 	{
		 		flag = 1;
		 		s[++tot].l = i;
		 	}
		 	if (!dp[n][i] && flag)
		 	{
		 		s[tot].r = i;
		 		flag = 0;
		 	}
		}
 	}
}
inline void prin()
{
 	int ans = 0;
 	for (int j = 1; j <= m; j++)
 		if (vis[j])
 			ans++;
 	if (ans != m)
 		printf("0\n%d", m - ans);
 	else
 	{
 		sort(s + 1, s + 1 + tot, cmp);
// 		for (int i = 1; i <= tot; i++)
// 			printf("%d %d\n", s[i].l, s[i].r);
// 		return;
 		int ans = 0, left = 0, right = 0;
	 	for (int i = 1; i <= tot; i++)
	 	{
			if (left >= s[i].l)//要加=号，这是贪心的线段覆盖的模板。
			right = max(right, s[i].r);
			else
		 	{
				ans++;
				left = right;
				right = max(right, s[i].r);
		 	}
			if (right > m) break;
		}
 		printf("1\n%d", ans);
 	}
}
int main()
{
	init();
 	bfs();
 	prin();
 	return 0;
}