HDU 5009 Paint Pearls（西安網路賽C題） dp+離散化+最佳化

標籤：des   style   blog   http   color   io   os   java   ar   

轉自：http://blog.csdn.net/accelerator_/article/details/39271751

吐血ac。。。

11668627

2014-09-16 22:15:24

Accepted

5009

1265MS

1980K

2290 B

G++

czy

 

Paint PearlsTime 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 k2 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 ≤ 5×104), indicating the number of pearls. The second line contains a1,a2,...,an (1 ≤ ai ≤ 109) indicating the target color of each pearl. Output   For each test case, output the minimal cost in a line. Sample Input31 3 3103 4 2 4 4 2 4 3 2 2 Sample Output27 Source 2014 ACM/ICPC Asia Regional Xi‘an Online  Recommendhujie   |   We have carefully selected several similar problems for you:  5017 5016 5014 5013 5011

 

轉自：http://blog.csdn.net/accelerator_/article/details/39271751

題意：給定一個目標顏色，每次能選一個區間染色，染色的代價為這個區間不同顏色數的平方，問最小代價

思路：先預先處理，把相同顏色的一段合并成一個點，然後把顏色離散化掉，然後進行dp，dp[i]表示染到第i個位置的代價，然後往後轉移，轉移的過程記錄下不同個數，這樣就可以轉移了，注意加個剪枝，就是如果答案大於了dp[n]就不用往後繼續轉移了

 

哎，dp思路還是很混亂，有空還要把這題好好做做。。。

 

  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 }

 

HDU 5009 Paint Pearls（西安網路賽C題） dp+離散化+最佳化