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

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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：

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#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; }