U187635 刷墙(easy)

题目描述：

n面墙，每面墙都有一个颜色，k个工人，每个工人至少刷一面墙，每个工人不是白干活，需要支付他工资，工资等于每个极大颜色段大小的平方和，即每个连续相同颜色块的长度的平方和，比如刷的是3 3 4 5 5 5，那工资就是2² + 4 ¹ + 5 ³，问刷完 n 面墙的最小花费是什么

思路：

区间dp

dp[i] [j] 表示 j 个工人刷完前 i 个墙的最小花费

$dp[i][j]=min(dp[p][j-1]+cost[p+1][i],dp[i][j]),j-1<=p<=i-1$

cost[l] [r]表示一个人刷 l 到 r 的区间需要的工资

然后类似于区间dp枚举len一样去枚举k，来更新dp[i] [j]

cost数组需要进行预处理，处理的时候枚举左端点和右端点，计算就行

不过还有有很多的细节需要注意，比如初始化之类的

/*
Work by: Chelsea
Problem: 不知名屑题
Knowledge: 垃圾算法
Time: O(能过)
*/
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<bitset>
#include<cstdio>
#include<string>
#include<vector>
#include<sstream>
#include<cstring>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<unordered_map>
using namespace std;

//#pragma GCC optimize("Ofast")
//#pragma GCC target("fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
//#pragma GCC optimize("unroll-loops")

#define eps 1e-8
#define endl '\n'
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define NMAX 1000 + 50
#define ls p<<1
#define rs p<<1|1
#define mod 1000000007
#define lowbit(x) (x & (-x))
#define sz(x) (int)(x).size()
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define sl(n) scanf("%lld",&n)
#define sll(n,m) scanf("%lld %lld",&n,&m)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d\n",n, m)
#define sddd(n,m,z) scanf("%d %d %d",&n,&m,&z)
#define slll(n,m,z) scanf("%lld %lld %lld",&n,&m,&z)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define mem(a,b) memset((a),(b),sizeof(a))
#define m_p(a,b) make_pair(a, b)
typedef  long long ll;
typedef pair <int,int> pii;
typedef unsigned long long ull;
//不开longlong见祖宗!
inline int IntRead(){char ch = getchar();int s = 0, w = 1;while(ch < '0' || ch > '9'){if(ch == '-') w = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){s = s * 10 + ch - '0';ch = getchar();}return s * w;}

//不改范围见祖宗!!!
#define MAX 1000000 + 50
int n, m, k;
ll tr[MAX];
ll cost[2005][2005];
ll dp[2005][35];
void work(){
    cin>>n>>k;
    for(int i = 1; i <= n; ++i){
        cin>>tr[i];
    }
    ll len = 1;
    for(int i = 1; i <= n; ++i)cost[i][i] = 1;
    for(int l = 1; l <= n; ++l){
        len = 1;
        for(int r = l + 1; r <= n; ++r){
            if(tr[r] == tr[r - 1]){
                cost[l][r] = cost[l][r - 1] + 2 * len + 1;
                ++len;
            }
            else{
                cost[l][r] = cost[l][r - 1] + 1;
                len = 1;
            }
        }
    }
    mem(dp, inf);
    for(int i = 1; i <= n; ++i)dp[i][1] = cost[1][i];
    for(int i = 1; i <= n; ++i){
        for(int j = 2; j <= min(i, k); ++j){
            for(int p = j - 1; p < i; ++p){
                dp[i][j] = min(dp[i][j], dp[p][j - 1] + cost[p + 1][i]);
            }
        }
    }
    cout<<dp[n][k]<<endl;
}

int main(){
    io;
//    int tt;cin>>tt;
//    for(int _t = 1; _t <= tt; ++_t){
//        printf("Case #%d: ", _t);
        work();
//    }
    return 0;
}