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 }。

//STATUS:C++_AC_0MS_276KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=110;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int f[N][N],sum[N],num[N];
int T,n;

int main()
{
 //  freopen("in.txt","r",stdin);
    int i,j,w,k,ca=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum[0]=0;
        for(i=1;i<=n;i++){
            scanf("%d",&num[i]);
            sum[i]=sum[i-1]+num[i];
        }
        for(i=1;i<=n;i++){
            for(j=i;j<=n;j++)
                f[i][j]=INF;
        }
        for(w=1;w<=n;w++){
            for(i=1;i+w-1<=n;i++){
                j=i+w-1;
                for(k=1;k<=w;k++){
                    f[i][j]=Min(f[i][j],f[i+1][i+k-1]+(k-1)*num[i]
                                +f[i+k][j]+k*(sum[j]-sum[i+k-1]));
                }
            }
        }

        printf("Case #%d: %d
",ca++,f[1][n]);
    }
    return 0;
}