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;
}