Code Forces 711C Coloring Trees
C. Coloring Trees
2 seconds
256 megabytes
standard input
standard output
ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the park where n trees grow. They decided to be naughty and color the trees in the park. The trees are numbered with integers from 1 to n from left to right.
Initially, tree i has color ci. ZS the Coder and Chris the Baboon recognizes only m different colors, so 0 ≤ ci ≤ m, where ci = 0 means that tree i is uncolored.
ZS the Coder and Chris the Baboon decides to color only the uncolored trees, i.e. the trees with ci = 0. They can color each of them them in any of the m colors from 1 to m. Coloring the i-th tree with color j requires exactly pi, j litres of paint.
The two friends define the beauty of a coloring of the trees as the minimum number of contiguous groups (each group contains some subsegment of trees) you can split all the n trees into so that each group contains trees of the same color. For example, if the colors of the trees from left to right are 2, 1, 1, 1, 3, 2, 2, 3, 1, 3, the beauty of the coloring is 7, since we can partition the trees into 7 contiguous groups of the same color : {2}, {1, 1, 1}, {3}, {2, 2}, {3}, {1}, {3}.
ZS the Coder and Chris the Baboon wants to color all uncolored trees so that the beauty of the coloring is exactly k. They need your help to determine the minimum amount of paint (in litres) needed to finish the job.
Please note that the friends can't color the trees that are already colored.
The first line contains three integers, n, m and k (1 ≤ k ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of trees, number of colors and beauty of the resulting coloring respectively.
The second line contains n integers c1, c2, ..., cn (0 ≤ ci ≤ m), the initial colors of the trees. ci equals to 0 if the tree number i is uncolored, otherwise the i-th tree has color ci.
Then n lines follow. Each of them contains m integers. The j-th number on the i-th of them line denotes pi, j (1 ≤ pi, j ≤ 109) — the amount of litres the friends need to color i-th tree with color j. pi, j's are specified even for the initially colored trees, but such trees still can't be colored.
Print a single integer, the minimum amount of paint needed to color the trees. If there are no valid tree colorings of beauty k, print - 1.
3 2 2
0 0 0
1 2
3 4
5 6
10
3 2 2
2 1 2
1 3
2 4
3 5
-1
3 2 2
2 0 0
1 3
2 4
3 5
5
3 2 3
2 1 2
1 3
2 4
3 5
0
In the first sample case, coloring the trees with colors 2, 1, 1 minimizes the amount of paint used, which equals to 2 + 3 + 5 = 10. Note that 1, 1, 1 would not be valid because the beauty of such coloring equals to 1 ({1, 1, 1} is a way to group the trees into a single group of the same color).
In the second sample case, all the trees are colored, but the beauty of the coloring is 3, so there is no valid coloring, and the answer is - 1.
In the last sample case, all the trees are colored and the beauty of the coloring matches k, so no paint is used and the answer is 0.
题意:给定N棵树,每棵树都有不同的颜色0-n,0代表没有涂色。如果这棵树没有颜色,那么你可以给他上色,如果已经有颜色,那你不能进行任何操作。接着给定一个数K,K代表着一排树的颜色有多少个相同的连续段。最后,给定每棵 树染上每种颜色的花费,求满足K的最小花费?
思路:比较明显的dp,但是当时并不会做,后来问了ZK大佬,才习得姿势,dp一直都不会,慢慢积累。
首先dp[i][j][k]代表,前i棵树的最后一棵树涂第j种颜色,并且涂成k个相同连续段的最小花费。首先初始状态dp[0][0][0]=0,然后分两种情况走。
1.color[i+1]!=0时,那么有2种情况:当前的j==color[i+1]时,dp[i+1][color[i+1]][k]=min(dp[i+1][color[i+1]][k],dp[i][j][k]),
当前的j!=color[i+1]时,dp[i+1][color[i+1]][k+1]=min(dp[i+1][color[i+1]][k+1],dp[i][j][k])。
2.color[i+1]==0时,那这棵树有1-p(m种)颜色可以选择,所以当p==j时,dp[i+1][p][k]=min(dp[i][j][k]+val[i+1][p],dp[i+1][p][k]),
当p!=j时,dp[i+1][p][k+1]=min(dp[i][j][k]+val[i+1][p],dp[i+1][p][k+1])。
代码如下:
#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <sstream>
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
#define mod 1000000007
#define mt(A,B) memset(A,B,sizeof(A))
using namespace std;
typedef long long LL;
const int N=200000+10;
const LL INF=0x3f3f3f3f3f3f3f3fLL;
LL dp[105][105][105];
LL c[105],val[105][105],ans=INF;
int main()
{
#ifdef Local
freopen("data.txt","r",stdin);
#endif
int i,j,K,n,m,k;
cin>>n>>m>>K;
for(i=1;i<=n;i++)cin>>c[i];
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
cin>>val[i][j];
}
}
mt(dp,0x3f);
dp[0][0][0]=0;
for(i=0;i<n;i++){
for(j=0;j<=m;j++){
for(k=0;k<=i;k++){
if(dp[i][j][k]!=INF){
if(c[i+1]){
dp[i+1][c[i+1]][k+(c[i+1]!=j)]=min(dp[i+1][c[i+1]][k+(c[i+1]!=j)],dp[i][j][k]);
}
else{
for(int p=1;p<=m;p++){
dp[i+1][p][k+(p!=j)]=min(dp[i][j][k]+val[i+1][p],dp[i+1][p][k+(p!=j)]);
}
}
}
}
}
}
for(i=1;i<=m;i++)
{
ans=min(dp[n][i][K],ans);
}
if(ans==INF)cout<<-1<<endl;
else cout<<ans<<endl; }
Code Forces 711C Coloring Trees的更多相关文章
- codeforces 711C Coloring Trees(DP)
题目链接:http://codeforces.com/problemset/problem/711/C O(n^4)的复杂度,以为会超时的 思路:dp[i][j][k]表示第i棵数用颜色k涂完后bea ...
- 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 ...
- 【动态规划】Codeforces 711C Coloring Trees
题目链接: http://codeforces.com/problemset/problem/711/C 题目大意: 给N棵树,M种颜色,已经有颜色的不能涂色,没颜色为0,可以涂色,每棵树I涂成颜色J ...
- CodeForces 711C Coloring Trees
简单$dp$. $dp[i][j][k]$表示:前$i$个位置染完色,第$i$个位置染的是$j$这种颜色,前$i$个位置分成了$k$组的最小花费.总复杂度$O({n^4})$. #pragma com ...
- Coloring Trees CodeForces - 711C
Coloring Trees CodeForces - 711C 题意:有n个点,每个点有一个c值,如果为0表示它没有被染色,否则表示它被染成了c值的颜色.颜色有1到m.把第i棵树染成颜色j所需要的代 ...
- codeforces 711C C. Coloring Trees(dp)
题目链接: C. Coloring Trees time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
- Codeforces Round #369 (Div. 2) C. Coloring Trees(dp)
Coloring Trees Problem Description: ZS the Coder and Chris the Baboon has arrived at Udayland! They ...
- Codeforces Round #369 (Div. 2) C. Coloring Trees DP
C. Coloring Trees ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the pa ...
- CodeForces #369 C. Coloring Trees DP
题目链接:C. Coloring Trees 题意:给出n棵树的颜色,有些树被染了,有些没有.现在让你把没被染色的树染色.使得beauty = k.问,最少使用的颜料是多少. K:连续的颜色为一组 ...
随机推荐
- 【转】Oracle job procedure 存储过程定时任务
原文:Oracle job procedure 存储过程定时任务 oracle job有定时执行的功能,可以在指定的时间点或每天的某个时间点自行执行任务. 一.查询系统中的job,可以查询视图 --相 ...
- 小笔记(三):PHP使用thinkphp3.2.3对数组进行分页
之前写过thinkphp3.2.3直接在查询数据的时候进行分页,前段时间用到了将查询之后的数组进行整理后进行分页,用到的一个函数array_slice($arr, $start, $length,tr ...
- FastCgi与PHP-fpm关系[转] 读完本文瞬间明朗了很多
刚开始对这个问题我也挺纠结的,看了<HTTP权威指南>后,感觉清晰了不少. 首先,CGI是干嘛的?CGI是为了保证web server传递过来的数据是标准格式的,方便CGI程序的编写者. ...
- 跨域、sql注入、xss攻击
这几天遇到这三个问题,现在简单的记录下来. 1.跨域 如我服务器的域名是www.test1.com,我在另一个服务器www.test2.com通过ajax访问www.test1.com的数据时,就引起 ...
- redis的安装-windows和linux
windows 下载地址:http://code.google.com/p/servicestack/wiki/RedisWindowsDownload 下载解压到D盘下: 进到该目录下,有下列文件: ...
- poj 3185 The Water Bowls
The Water Bowls 题意:给定20个01串(最终的状态),每个点变化时会影响左右点,问最终是20个0所需最少操作数? 水题..直接修改增广矩阵即可:看来最优解不是用高斯消元(若是有Gaus ...
- [简历] JAVA 软件工程师
首先,一份好的简历不光说明事实,更通过FAB模式来增强其说服力. Feature:是什么 Advantage:比别人好在哪些地方 Benefit:如果雇佣你,招聘方会得到什么好处 其次,写简历和写议论 ...
- UITextView -- 基础备忘
UITextView 这篇文章只涉及到基本的使用,日后会写一些关于结合TextKit的备忘 基本属性 let screenSize = UIScreen.mainScreen().bounds.siz ...
- Python读写Redis数据库
import redis class Database: def __init__(self): self.host = 'localhost' self.port = 6379 def write( ...
- 【HDOJ】1512 Monkey King
左偏树+并查集.左偏树就是可合并二叉堆. /* 1512 */ #include <iostream> #include <string> #include <map&g ...