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POJ 1651 Multiplication Puzzle (区间dp)

时间:2014-08-05 03:01:28      阅读:217      评论:0      收藏:0      [点我收藏+]

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题目大意:对n个数组成的序列取数,规定最两边不能取,每次取一个a[i],得到 a[l] * a[i] * a[r] 的分数(a[l]是a[i]左边的数,a[r]是a[i]右边的数),并把这个数从序列中移走,求n-2次取数后的得分和的最小值

分析:正着确定状态不好做,不如反着来,设dp[l][r]为向区间[l, r]中填满数所得到分数和的最小值,考虑最近一次填数的位置,不难得出:

dp[l][r] = fmin(dp[l][m] + dp[m][r] + a[l] * a[m] * a[r]) (l<m<r);

实现方式可以选择记忆化递归,也可直接以区间长度为阶段来递推,各有优缺点,代码如下:

bubuko.com,布布扣
  1 /**********************************************
  2 ***    Problem:
  3 ***    Author:        JKL
  4 ***    University:    CSUST
  5 ***    Team:          __Dream
  6 ***    Email:          1451108308@QQ.COM
  7 ***    My Blog:        http://www.cnblogs.com/jklongint/
  8 ***********************************************/
  9 //===================================================
 10 #include <iostream>
 11 #include <fstream>
 12 #include <sstream>
 13 #include <iomanip>
 14 #include <cstdio>
 15 #include <cstdlib>
 16 #include <cmath>
 17 #include <cassert>
 18 #include <numeric>
 19 #include <ctime>
 20 #include <algorithm>
 21 #include <cstring>
 22 #include <string>
 23 #include <vector>
 24 #include <queue>
 25 #include <map>
 26 #include <stack>
 27 #include <list>
 28 #include <set>
 29 #include <bitset>
 30 #include <deque>
 31 using namespace std;
 32 //---------------------------------------------------
 33 #define mem(a,b) memset(a,b,sizeof(a))
 34 #define GO cout<<"HelloWorld!"<<endl
 35 #define Case(x) cout<<"Case "<<x<<":"
 36 #define foru(i,n) for(int i=1; i <= n; i++)
 37 #define ford(i,n) for(int i = n; i >= 1; i--)
 38  #define fin freopen("input.txt","r",stdin);
 39  #define fout freopen("output.txt","w",stdout)
 40 #define lson  l, m, rt << 1
 41 #define rson  m + 1, r, rt << 1 | 1
 42 
 43 #define sqr(a)  ((a)*(a))
 44 #define abs(a) ((a>0)?(a):-(a))
 45 #define pii pair<int,int>
 46 
 47 #define fmax(a,b) max(a,b)
 48 #define fmin(a,b) min(a,b)
 49 #define fmax3(a,b,c)  (fmax(a,fmax(a,b)))
 50 #define fmin3(a,b,c)  (fmin(a,fmin(a,b)))
 51 
 52 #define sfi(x) scanf("%d",&x)
 53 #define sfL(x) scanf("%I64d",&x)
 54 #define sfc(x) scanf("%c",&x)
 55 #define sfd(x) scanf("%lf",&x)
 56 #define sfs(x) scanf("%s",x)
 57 #define sfii(a,b) scanf("%d%d",&a,&b)
 58 #define sfLL(a,b) scanf("%I64d%I64d",&a,&b)
 59 #define sfcc(a,b) scanf("%c%c",&a,&b)
 60 #define sfdd(a,b) scanf("%lf%lf",&a,&b)
 61 #define sfss(a,b) scanf("%s%s",a,b)
 62 
 63 #define pfi(x) printf("%d",x)
 64 #define pfL(x) printf("%I64d",x)
 65 #define pfs(x) printf("%s",x)
 66 #define pfd(x) printf("%lf",x)
 67 #define pfc(x) print("%c",x)
 68 #define newLine pfs("\n")
 69 #define space pfs(" ")
 70 
 71 //--------------------------------------------------------
 72 typedef long long LL;
 73 typedef unsigned long long ULL;
 74 typedef __int64 __LL;
 75 typedef unsigned __int64 __ULL;
 76 
 77 typedef vector<int> vi;
 78 typedef vector<LL> vL;
 79 typedef vector<string> vs;
 80 typedef set<int> si;
 81 typedef map<int,int> mii;
 82 typedef map<LL,LL> mLL;
 83 typedef map<string,int> msi;
 84 typedef map<char,int> mci;
 85 //--------------------------------------------------------
 86 const int dx[4]={1,-1,0,0};
 87 const int dy[4]={0,0,1,-1};
 88  const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 89  const int N6=1000006;
 90  const int N5=100006;
 91  const int N4=10006;
 92  const int N3=1006;
 93  const int N2=106;
 94  const int N=1009;
 95  const int MOD=1000000007;
 96  const LL LMAX=0x7fffffffffffffff;
 97  const LL IMAX=0x3fffffff;
 98  const double PI=3.14159265359;
 99 //--------------------------------------------------------
100 template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
101 template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }
102 
103 //------------------------------------------------------------
104 struct TreeNode{
105     LL  sum, add;
106 };
107 //=================================================================
108 TreeNode tree[N << 3];
109 int flag[N][N], a[N], dp[N][N];
110 int dpSearch(int l, int r)
111 {
112     if(l + 1 == r)return 0;
113     if(flag[l][r])return dp[l][r];
114     dp[l][r] = IMAX;
115     for(int j = l + 1; j < r; j++){
116         int s = dpSearch(l, j), t = dpSearch(j, r);
117         //if(l == 1 && r == 4)cout<< s<< t<< endl;
118         dp[l][r] = fmin(dp[l][r], s + t + a[l] * a[j] * a[r]);
119     }
120     flag[l][r] = 1;
121     //cout<<l<<r<<dp[l][r]<<endl;
122     return dp[l][r];
123 }
124 int main()
125 {
126     //fin;//fout;//freopen("input.txt","r",stdin);
127     int n;
128     cin>>n;
129     foru(i, n)sfi(a[i]);
130     cout << dpSearch(1, n) << endl;
131     return 0;
132 }
View Code

 

POJ 1651 Multiplication Puzzle (区间dp),布布扣,bubuko.com

POJ 1651 Multiplication Puzzle (区间dp)

标签:style   blog   http   color   os   io   for   art   

原文地址:http://www.cnblogs.com/jklongint/p/3891346.html

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