CodeForces 711C Coloring Trees (DP)
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;
- }
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