Codeforces Round #369 (Div. 2) C 四维dp

题目传送门 ： http://codeforces.com/contest/711/problem/C

题意：有n颗树，m中颜料，对树进行涂色，涂成k组（如1，2，1为三组；1，2，2为两组；1，1，1为一组）。接下里一行n个数，0代表未涂色，正数代表已涂色。接下来n行表示 第i棵树染成j色需要Pi,j吨颜料。问想要符合条件，最少需要耗费多少吨颜料

思路：三维dp，空间三维，时间n^4（CF机器果然快，1s差不多常熟倍的10^8），第一维i是树的个数，二维j是颜色的组数，三维l是i-1涂的颜色，状态转移方程，分为四个，天生带颜色的和不带颜色的各两个

if(c[i] > 0){          //天生带颜色的
    if(c[i] == l) dp[i][j][c[i]] = min(dp[i][j][c[i]],dp[i-1][j][l]);
    else dp[i][j][c[i]] = min(dp[i][j][c[i]],dp[i-1][j-1][l]);
}
else{                  //需要进行涂色的
    for(int q=1;q<=m;q++){
        if(q == l) dp[i][j][q] = min(dp[i][j][q],dp[i-1][j][l] + p[i][q]);
        else dp[i][j][q] = min(dp[i][j][q],dp[i-1][j-1][l] + p[i][q]);
    }
}

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>

using namespace std;
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   ll            long long
#define   ull           unsigned long long
#define   mem(n,v)      memset(n,v,sizeof(n))
#define   MAX           105
#define   MAXN          3000
#define   PI            3.1415926
#define   E             2.718281828459
#define   opnin         freopen("text.in.txt","r",stdin)
#define   opnout        freopen("text.out.txt","w",stdout)
#define   clsin         fclose(stdin)
#define   clsout        fclose(stdout)
#define   haha1         cout << "haha1"<< endl
#define   haha2         cout << "haha2"<< endl
#define   haha3         cout << "haha3"<< endl

const int    INF    =   0x3f3f3f3f;
const ll     INFF   =   0x3f3f3f3f3f3f3f3f;
const double pi     =   3.141592653589793;
const double inf    =   1e18;
const double eps    =   1e-8;
const ll     mod    =   1e9+7;
const ull    mx     =   133333331;

ll c[MAX];
ll p[MAX][MAX];
ll dp[MAX][MAX][MAX];
int main()
{
    mem(c,0);
    mem(p,0);
    mem(dp,INFF);
    ll n,m,k;
    cin >> n >> m >> k;
    for(int i=1;i<=n;i++)
        scanf("%I64d",&c[i]);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%I64d",&p[i][j]);
        }
    }
    if(c[1] > 0)
        dp[1][1][c[1]] = 0;
    else{
        for(int j=1;j<=m;j++){
            dp[1][1][j] = p[1][j];
        }
    }
    for(int i=2;i<=n;i++){
        for(int j=1;j<=n && j<=k;j++){
            for(int l=1;l<=m;l++){
                if(c[i] > 0){
                    if(c[i] == l) dp[i][j][c[i]] = min(dp[i][j][c[i]],dp[i-1][j][l]);
                    else dp[i][j][c[i]] = min(dp[i][j][c[i]],dp[i-1][j-1][l]);
                }
                else{
                    for(int q=1;q<=m;q++){
                        if(q == l) dp[i][j][q] = min(dp[i][j][q],dp[i-1][j][l] + p[i][q]);
                        else dp[i][j][q] = min(dp[i][j][q],dp[i-1][j-1][l] + p[i][q]);
                    }
                }
            }
        }
    }
    ll Min = INFF;
    for(int i=1;i<=m;i++)
        Min = min(dp[n][k][i],Min);
    if(Min != INFF) cout << Min <<endl;
    else cout << "-1" << endl;
    return 0;
}