HDU-4283 You Are the One 区间DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4283

　　题意：n个人排队，每个人有一个权值val[i]。从第一个人开始出队，进入一个栈中，每次可以留在栈中或者从栈中移出一个，如果第 i 个人是第k个出栈的，那么有sum+=(k-1)*val[i]，求是的sum最小。

　　f[i][j]表示区间第 i 个人到第 j 个人sum的最小值，那么每次转移的时候我们只要枚举第 i 个人是什么时候出的栈就可以了，假设第 i 个人是第k个出的栈，那么f[i][j]=Min{ f[i+1][i+k-1]+(k-1)*num[i]+f[i+k][j]+k*(sum[j]-sum[i+k-1]) | 1<=k<=j-i+1 }。

 1 //STATUS:C++_AC_0MS_276KB

 2 #include <functional>

 3 #include <algorithm>

 4 #include <iostream>

 5 //#include <ext/rope>

 6 #include <fstream>

 7 #include <sstream>

 8 #include <iomanip>

 9 #include <numeric>

10 #include <cstring>

11 #include <cassert>

12 #include <cstdio>

13 #include <string>

14 #include <vector>

15 #include <bitset>

16 #include <queue>

17 #include <stack>

18 #include <cmath>

19 #include <ctime>

20 #include <list>

21 #include <set>

22 #include <map>

23 using namespace std;

24 //#pragma comment(linker,"/STACK:102400000,102400000")

25 //using namespace __gnu_cxx;

26 //define

27 #define pii pair<int,int>

28 #define mem(a,b) memset(a,b,sizeof(a))

29 #define lson l,mid,rt<<1

30 #define rson mid+1,r,rt<<1|1

31 #define PI acos(-1.0)

32 //typedef

33 typedef __int64 LL;

34 typedef unsigned __int64 ULL;

35 //const

36 const int N=110;

37 const int INF=0x3f3f3f3f;

38 const int MOD=100000,STA=8000010;

39 const LL LNF=1LL<<60;

40 const double EPS=1e-8;

41 const double OO=1e15;

42 const int dx[4]={-1,0,1,0};

43 const int dy[4]={0,1,0,-1};

44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

45 //Daily Use ...

46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

56 //End

57 

58 int f[N][N],sum[N],num[N];

59 int T,n;

60 

61 int main()

62 {

63  //  freopen("in.txt","r",stdin);

64     int i,j,w,k,ca=1;

65     scanf("%d",&T);

66     while(T--)

67     {

68         scanf("%d",&n);

69         sum[0]=0;

70         for(i=1;i<=n;i++){

71             scanf("%d",&num[i]);

72             sum[i]=sum[i-1]+num[i];

73         }

74         for(i=1;i<=n;i++){

75             for(j=i;j<=n;j++)

76                 f[i][j]=INF;

77         }

78         for(w=1;w<=n;w++){

79             for(i=1;i+w-1<=n;i++){

80                 j=i+w-1;

81                 for(k=1;k<=w;k++){

82                     f[i][j]=Min(f[i][j],f[i+1][i+k-1]+(k-1)*num[i]

83                                 +f[i+k][j]+k*(sum[j]-sum[i+k-1]));

84                 }

85             }

86         }

87 

88         printf("Case #%d: %d\n",ca++,f[1][n]);

89     }

90     return 0;

91 }