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CodeForces 711C Coloring Trees (DP)

目录

作者：@dwtfukgv
本文为作者原创，转载请注明出处：https://www.cnblogs.com/dwtfukgv/p/5824235.html

题意：给定n棵树，其中有一些已经涂了颜色，然后让你把没有涂色的树涂色使得所有的树能够恰好分成k组，让你求最少的花费是多少。

析：这是一个DP题，dp[i][j][k]表示第 i 棵树涂第 j 种颜色恰好分成 k 组，然后状态转移方程是什么呢？

如果第 i 棵已经涂了，那么要么和第 i-1 棵一组，要么不和第 i-1 棵一组。

如果第 i 棵没有涂，和上面差不多，就是加上要涂的费用，并且要选择最少的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <list>
#include <sstream>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL cor[maxn], w[maxn][maxn];
LL dp[maxn][maxn][maxn];
 
int main(){
    int K;
    while(scanf("%d %d %d", &n, &m, &K) == 3){
        for(int i = 1; i <= n; ++i)  scanf("%I64d", &cor[i]);
        for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j)  scanf("%I64d", &w[i][j]);
 
        for(int i = 0; i <= n; ++i)  for(int j = 0; j <= m; ++j)
            for(int k = 0; k <= K; ++k)  dp[i][j][k] = LNF;
 
        dp[0][0][0] = 0;
        for(int i = 1; i <= n; ++i){
            if(cor[i]){
                for(int k = 1; k <= K; ++k){
                    dp[i][cor[i]][k] = Min(dp[i][cor[i]][k], dp[i-1][cor[i]][k]);
                    for(int j = 0; j <= m; ++j){
                        if(j != cor[i])  dp[i][cor[i]][k] = Min(dp[i][cor[i]][k], dp[i-1][j][k-1]);
                    }
                }
            }
            else{
                for(int k = 1; k <= K; ++k){
                    for(int j = 1; j <= m; ++j){
                        dp[i][j][k] = Min(dp[i][j][k], dp[i-1][j][k] + w[i][j]);
                        for(int l = 0; l <= m; ++l){
                            if(l != j)  dp[i][j][k] = Min(dp[i][j][k], dp[i-1][l][k-1] + w[i][j]);
                        }
                    }
                }
            }
        }
        LL ans = LNF;
        for(int i = 1; i <= m; ++i)  ans = Min(ans, dp[n][i][K]);
        if(ans == LNF)  ans = -1;
        cout << ans << endl;
    }
    return 0;
}