HDU 5009 Paint Pearls（西安网络赛C题） dp+离散化+优化

最新推荐文章于 2019-08-27 15:59:48 发布

转载最新推荐文章于 2019-08-27 15:59:48 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/njczy2010/p/3975968.html

本文探讨了一个关于优化染色方案的问题，通过动态规划的方法求解最小代价。详细介绍了预处理、颜色离散化、DP状态转移等关键步骤，并提供了具体的代码实现和示例输入输出。

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转自：http://blog.youkuaiyun.com/accelerator_/article/details/39271751

吐血ac。。。

11668627

2014-09-16 22:15:24

Accepted

5009

1265MS

1980K

2290 B

G++

czy

Paint Pearls

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1473 Accepted Submission(s): 466

Problem Description

   Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help.
   In each operation, he selects some continuous pearls and all these pearls will be painted to their target colors. When he paints a string which has k different target colors, Lee will cost k ² points.
   Now, Lee wants to cost as few as possible to get his ideal string. You should tell him the minimal cost.

Input

There are multiple test cases. Please process till EOF.
For each test case, the first line contains an integer n(1 ≤ n ≤ 5×10 ⁴), indicating the number of pearls. The second line contains a ₁,a ₂,...,a _n (1 ≤ a _i ≤ 10 ⁹) indicating the target color of each pearl.

Output

For each test case, output the minimal cost in a line.

Sample Input

       3 1 3 3 10 3 4 2 4 4 2 4 3 2 2
      

Sample Output

2 7

Source

2014 ACM/ICPC Asia Regional Xi'an Online

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转自：http://blog.youkuaiyun.com/accelerator_/article/details/39271751

题意：给定一个目标颜色，每次能选一个区间染色，染色的代价为这个区间不同颜色数的平方，问最小代价

思路：先预处理，把相同颜色的一段合并成一个点，然后把颜色离散化掉，然后进行dp，dp[i]表示染到第i个位置的代价，然后往后转移，转移的过程记录下不同个数，这样就可以转移了，注意加个剪枝，就是如果答案大于了dp[n]就不用往后继续转移了

哎，dp思路还是很混乱，有空还要把这题好好做做。。。

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<string>
 10 
 11 #define N 50005
 12 #define M 15
 13 #define mod 10000007
 14 #define p 10000007
 15 #define mod2 100000000
 16 #define ll long long
 17 #define LL long long
 18 #define maxi(a,b) (a)>(b)? (a) : (b)
 19 #define mini(a,b) (a)<(b)? (a) : (b)
 20 
 21 using namespace std;
 22 
 23 int n,k,s;
 24 int a[N];
 25 int b[N];
 26 map<int,int>c;
 27 int vis[N];
 28 int dp[N];
 29 int cou;
 30 vector<int>save;
 31 
 32 void ini()
 33 {
 34     //memset(vis,0,sizeof(vis));
 35     memset(dp,0x3f3f3f3f,sizeof(dp));
 36     c.clear();
 37     k=0;
 38     int i;
 39     scanf("%d",&a[1]);
 40     k=1;
 41     b[1]=a[1];
 42     for(i=2;i<=n;i++){
 43        scanf("%d",&a[i]);
 44        if(a[i]!=a[i-1]){
 45           k++;
 46           b[k]=a[i];
 47        }
 48     }
 49     s=0;
 50     for(i=1;i<=k;i++){
 51        if(c[ b[i] ]==0){
 52          // vis[ b[i] ]=1;
 53           s++;
 54           c[ b[i] ]=s;
 55        }
 56     }
 57 
 58     for(i=1;i<=k;i++){
 59        b[i]=c[ b[i] ];
 60       // dp[i]=i;
 61     }
 62    // for(i=1;i<=k;i++){
 63    //    printf(" i=%d b=%d\n",i,b[i]);
 64     //}
 65 
 66 }
 67 
 68 void solve()
 69 {
 70     int i,j;
 71     dp[0]=0;
 72     dp[k]=k;
 73     for(i=0;i<k;i++){
 74         cou=0;
 75        // vis[ b[i] ]=1;
 76         //save.push_back(b[i]);
 77         for(j=i+1;j<=k;j++){
 78            // if(cou*cou>=k) break;
 79             if(vis[ b[j] ]==0 ){
 80                 vis[ b[j] ]=1;
 81                 save.push_back(b[j]);
 82                 cou++;
 83             }
 84             if (dp[i] + cou * cou >= dp[k]) break;
 85            // printf("  i=%d j=%d dpj=%d cou=%d dp=%d ",i,j,dp[j],cou,dp[i]+cou*cou);
 86             dp[j]=min(dp[j],dp[i]+cou*cou);
 87            // printf("   dpj=%d\n",dp[j]);
 88         }
 89         for(vector<int>::iterator it=save.begin();it!=save.end();it++){
 90             vis[*it]=0;
 91         }
 92         save.clear();
 93     }
 94 }
 95 
 96 void out()
 97 {
 98     //for(int i=1;i<=k;i++){
 99     //    printf(" i=%d dp=%d\n",i,dp[i]);
100     //}
101     printf("%d\n",dp[k]);
102 }
103 
104 int main()
105 {
106     //freopen("data.in","r",stdin);
107     //freopen("data.out","w",stdout);
108     //scanf("%d",&T);
109     //for(int cnt=1;cnt<=T;cnt++)
110    // while(T--)
111     while(scanf("%d",&n)!=EOF)
112     {
113         ini();
114         solve();
115         out();
116     }
117 
118     return 0;
119 }

转载于:https://www.cnblogs.com/njczy2010/p/3975968.html