CF1114D Flood Fill

题目

题目描述

You are given a line of nn colored squares in a row, numbered from 11 to nnn from left to right. The iii -th square initially has the color cic_ici .Let’s say, that two squares iii and jjj belong to the same connected component if ci=cjc_i = c_jci=cj , and ci=ckc_i = c_kci=ck for all kkk satisfying i<k<ji < k < ji<k<j . In other words, all squares on the segment from iii to jjj should have the same color.For example, the line [3,3,3][3, 3, 3][3,3,3] has 11 connected component, while the line [5,2,4,4][5, 2, 4, 4][5,2,4,4] has 333 connected components.The game “flood fill” is played on the given line as follows:At the start of the game you pick any starting square (this is not counted as a turn).Then, in each game turn, change the color of the connected component containing the starting square to any other color.Find the minimum number of turns needed for the entire line to be changed into a single color.

输入格式

The first line contains a single integer nnn ( 1≤n≤50001\le n \le50001≤n≤5000 ) — the number of squares.The second line contains integers c1,c2,…,cnc_1, c_2, \ldots, c_nc1,c2,…,cn ( 1≤ci≤50001 \le c_i \le 50001≤ci≤5000 ) — the initial colors of the squares.

输出格式

Print a single integer — the minimum number of the turns needed.

题意翻译

n 个方块排成一排，第 iii 个颜色为 cic_ici。定义一个颜色联通块 [l,r][l,r][l,r] 当且仅当 lll 和 rrr 之间（包括 l,rl,rl,r）所有方块的颜色相同。现在你可以选定一个起始位置 ppp，每次将 ppp 所在颜色联通块的所有方块颜色改成另一种。这个操作可能将多个颜色联通块合并成一个。问最少要多少步，能让 [1,n][1,n][1,n]变成一个颜色联通块

数据范围

1≤n,ci≤5000。1≤n,ci≤5000。1≤n,ci≤5000。

输入输出样例

输入 #1

4
5 2 2 1

输出 #1

输入 #2

8
4 5 2 2 1 3 5 5

输出 #2

输入 #3

1
4

输出 #3

说明/提示

In the first example, a possible way to achieve an optimal answer is to pick square with index 222 as the starting square and then play as follows:

[5,2,2,1][5, 2, 2, 1][5,2,2,1]
[5,5,5,1][5, 5, 5, 1][5,5,5,1]
[1,1,1,1][1, 1, 1, 1][1,1,1,1]

In the second example, a possible way to achieve an optimal answer is to pick square with index 555 as the starting square and then perform recoloring into colors 2,3,5,42, 3, 5, 42,3,5,4 in that order.In the third example, the line already consists of one color only.

原题链接

题解

思路

这道题是说的选定一个位置，接着不停地改变这个位置所在颜色联通块的颜色，直到所有位置颜色都相同。这里注意选定的位置不能改变，只能在同一个地方进行改变颜色的操作。这道题的算法比较好分析，可以进行区间合并，那肯定就是区间dp了，但是看到数据范围， nnn最大是5000，一般的区间dp都是
O(n3)O(n^3)O(n3), 这是肯定会超时的，所以我们应该想办法省掉一层循环。最外层循环是枚举的区间长度，也就是区间dp的阶段，第二层循环是枚举的起始位置，这两个很明显都不能去掉，看来看去能够省下来的只能是最后枚举划分位置的那一层循环了。接下来我们就要想如何省了。我们通过读题，知道了进行改变颜色的位置只能是在同一处，所以我们枚举的每个两边不相等区间都可以由通过改变其中长度少1的区间变为剩下的那一个位置而得到，既是
dp[l][r]=min(dp[l+1][r]+1,dp[l][r−1]+1)dp[l][r] = min (dp[l + 1][r] + 1, dp[l][r - 1] + 1)dp[l][r]=min(dp[l+1][r]+1,dp[l][r−1]+1)
而当两头一样的话就只需要改变内部，使这个区间除去两端的区间颜色改变成两端一样的区间，既是
dp[l][r]=dp[l+1][r−1]+1dp[l][r] = dp[l + 1][r - 1] + 1dp[l][r]=dp[l+1][r−1]+1
不过此时又有问题了，如果里面还是一个两头相等且和外面一样的区间，这样就不能加1了，比如2,2,2,22, 2, 2, 22,2,2,2这里一次都不需要改，可是直接执行上面的方法就会输出222，这时候就需要在输入时就把所有颜色联通块的长度压缩成1。
方法：

for (int i = 1; i <= n; i++){scanf ("%d", &s[i]);if (s[i] == s[i - 1]){i --;n --;}
}

这样就可以避免上述情况发生

运用算法

区间dp区间dp区间dp

代码

#include <cstdio>
#include <algorithm>
using namespace std;#define MAXN 5000int s[MAXN + 5], dp[MAXN + 5][MAXN + 5];int main(){int n;scanf ("%d", &n);for (int i = 1; i <= n; i++){scanf ("%d", &s[i]);if (s[i] == s[i - 1]){//压缩联通块 i --;n --;}}//dp[i][i]就是0， 所以不用初始化 for (int i = 2; i <= n; i++){//枚举长度 for (int l = 1; l <= n - i + 1; l++){int r = l + i - 1;if (s[l] == s[r]){//如果两端相等， 就直接等于去除两端的区间加1 dp[l][r] = dp[l + 1][r - 1] + 1;}else {//否则比较从哪个变过来小 dp[l][r] = min (dp[l + 1][r], dp[l][r - 1]) + 1; }}}printf ("%d", dp[1][n]);//输出整个区间的变化次数 return 0;
}