HDU 4283 You Are the One (区间DP)
2017-08-12 20:29
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You Are the One
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4352 Accepted Submission(s): 2035
[align=left]Problem Description[/align]
The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot
of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness
of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow,
the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?
[align=left]Input[/align]
The first line contains a single integer T, the number of test cases. For each case, the first line is n (0 < n <= 100)
The next n line are n integer D1-Dn means the value of diaosi of boys (0 <= Di <= 100)
[align=left]Output[/align]
For each test case, output the least summary of unhappiness .
[align=left]Sample Input[/align]
2
5
1
2
3
4
5
5
5
4
3
2
2
[align=left]Sample Output[/align]
Case #1: 20
Case #2: 24
[align=left]Source[/align]
2012 ACM/ICPC Asia Regional Tianjin
Online
[align=left]Recommend[/align]
liuyiding
dp[i][j]表示第i个人到第j个人这段区间所需要的最小值
对于dp[i][j],只分析i到j这段区间,若第i个人是i到j这段区间里第k个上场的,
那么区间[i+1, i+k-1]这k-1个元素一定在i之前上台,区间[i+k, j]一定在i之后上台。
于是得转移方程:dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]+v[i]*(k-i)+(k-i+1)*(sum[j]-sum[k]))
#include <stdio.h> #include <string.h> #include <algorithm> const int INF=0x3f3f3f3f; using namespace std; int dp[110][110],v[110],sum[110],n; int main () { int t,cnt=0,i,j,k,l; scanf("%d",&t); while(t--) { cnt++; memset(dp,0,sizeof(dp)); memset(sum,0,sizeof(sum)); scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&v[i]); sum[i]=sum[i-1]+v[i]; } for(i=1;i<=n;i++) { for(j=i+1;j<=n;j++) dp[i][j]=INF; } for(l=1;l<n;l++) { for(i=1;i<=n-l;i++) { j=i+l; for(k=i;k<=j;k++) dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]+v[i]*(k-i)+(k-i+1)*(sum[j]-sum[k])); } } printf("Case #%d: %d\n",cnt,dp[1] ); } return 0; }
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