Codeforces Round #369 (Div. 2)-C. Coloring Trees DP

CF717C

题意：有长度为n的数列，有m种颜色，问最少的花费，使得数列中为0的点刷上颜色，并可根据颜色把数列分为正好k段。

思路：dp，开一个dp[n][m][k]，表示前n个以m为结尾的k段最小花费，转移方程是：如果这个点非0，那么只能从n-1 转移到a[i]这一个颜色，如果是0，即没有颜色限制，那么可以从n-1个中以非m结尾的状态转移过来，即dp[n][非m][k-1] .或者是以m为结尾的状态转移，即dp[n][m][k];

//#include<bits/stdc++.h>
//#include<unordered_map>
//#include<unordered_set>
#include<functional>
#include<algorithm>
#include<iostream>
#include<iomanip>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<cstddef>
#include<cstdio>
#include<memory>
#include<vector>
#include<cctype>
#include<string>
#include<cmath>
#include<queue>
#include<deque>
#include<ctime>
#include<stack>
#include<map>
#include<set>

#define fi first
#define se second
#define pb push_back
#define INF 0x3f3f3f3f
#define pi 3.1415926535898
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define Min(a,b,c)  min(a,min(b,c))
#define Max(a,b,c)  max(a,max(b,c))

// #pragma GCC optimize("unroll-loops")
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC optimize("Ofast,no-stack-protector")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;

typedef long long ll;
typedef pair<int,int> P;
typedef unsigned long long ull;

const int MOD=1e9+7;
const ll LLMAX=2e18;
const int MAXN=1e6+10;

template<class T>
inline void read(T &DataIn)
{
    DataIn=0;    T Flag=0;   char c=getchar();
    while(!isdigit(c)){ Flag|=c=='-'; c=getchar(); }
    while(isdigit(c)){ DataIn=DataIn*10+c-'0'; c=getchar(); }
    DataIn= Flag? -DataIn: DataIn;
}

template<class T>
inline void write(T DataOut,char EndChar='
')
{
    T lenth=0,number[30];
    if(DataOut==0){ putchar(48); return; }
    while(DataOut>0){ number[++lenth]=DataOut%10; DataOut/=10;}
    for(int i=lenth;i>=1;i--)    putchar(number[i]+48);
    putchar(EndChar);
}

priority_queue<int,vector<int>,less<int> > qd;
priority_queue<int,vector<int>,greater<int> > qu;

ll a[200][200],origin[200],change[200];

ll dp[200][200][200];


const ll inff = 0x3f3f3f3f3f3f3f3f; //18
int main(void)
{
    FILE *fin=NULL,*fout=NULL;
    ios::sync_with_stdio(false);    cin.tie(0);
    //fin=freopen("D:/Project__C++/testdata.in","r",stdin);
    //fout=freopen("D:/Project__C++/testdata.out","w",stdout);
    int n,m,k;  cin>>n>>m>>k;
    for(int i=1;i<=n;i++)   cin>>origin[i];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            cin>>a[i][j];
    
    for(int i=0; i<=n; i++)
        for(int j=0; j<=m; j++)
            for(int t = 0; t<=k;t++)
                dp[i][j][t] = inff;
            

    for(int i=1; i<=m; i++)dp[0][i][0]=0ll;
    dp[0][1][1] = 0;                        //这个初始条件是看到
                                            //1 1 1 0 5 这个样例发现的
    for(int i=1; i<=n; i++){

            for(int j=1; j<= m; j++){

                for(int f = 1; f<=m; f++){
                    for(int t = 1; t<=min(i,k); t++){       //这里要去min才行。

                        if(origin[i] != 0){

                             if(origin[i]==j){
                                if(j==f)   dp[i][j][t] = min(dp[i][j][t],dp[i-1][f][t]);
                                else dp[i][j][t] = min(dp[i-1][f][t-1],dp[i][j][t]);
                             }
                             
                        }
                        else {
                      
                            if(j!=f)dp[i][j][t] = min(dp[i-1][f][t-1] + a[i][j],dp[i][j][t]);
                            else dp[i][j][t] = min ( dp[i-1][f][t] + a[i][j],dp[i][j][t]);

                        }
                        // cout<<i<<" "<<j<<" "<<t<<":";
                        // cout<<dp[i][j][t]<<endl;
                    }
                }
            }
            // cout<<endl;
    }


    ll ans=inff;
    for(int i=1; i<=m; i++)  if(ans > dp[n][i][k]) ans = dp[n][i][k];
    if(ans>=inff)    ans=-1;
    cout<<ans<<endl;
    return 0;
}