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 }
