poj 1651 Multiplication Puzzle 【区间dp】
2015-12-24 22:30
411 查看
Multiplication Puzzle
Description
The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on
the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.
The goal is to take cards in such order as to minimize the total number of scored points.
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.
Input
The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.
Output
Output must contain a single integer - the minimal score.
Sample Input
Sample Output
题意:给你n个数num[],去掉第i个数(不能是第一个和最后一个)的代价 = num[i] * num[i-1] * num[i+1]。现在问你去掉中间n-2个数的最小代价。
思路:很基础的区间dp了。
枚举最后一个去掉的数。状态转移
dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k+1][j] + num[k] * num[i-1] * num[j+1]);
AC代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7926 | Accepted: 4892 |
The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on
the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.
The goal is to take cards in such order as to minimize the total number of scored points.
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.
Input
The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.
Output
Output must contain a single integer - the minimal score.
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
题意:给你n个数num[],去掉第i个数(不能是第一个和最后一个)的代价 = num[i] * num[i-1] * num[i+1]。现在问你去掉中间n-2个数的最小代价。
思路:很基础的区间dp了。
枚举最后一个去掉的数。状态转移
dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k+1][j] + num[k] * num[i-1] * num[j+1]);
AC代码:
#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #define INF 0x3f3f3f3f #define eps 1e-8 #define MAXN (100+10) #define MAXM (100000) #define Ri(a) scanf("%d", &a) #define Rl(a) scanf("%lld", &a) #define Rf(a) scanf("%lf", &a) #define Rs(a) scanf("%s", a) #define Pi(a) printf("%d\n", (a)) #define Pf(a) printf("%.2lf\n", (a)) #define Pl(a) printf("%lld\n", (a)) #define Ps(a) printf("%s\n", (a)) #define W(a) while(a--) #define CLR(a, b) memset(a, (b), sizeof(a)) #define MOD 1000000007 #define LL long long #define lson o<<1, l, mid #define rson o<<1|1, mid+1, r #define ll o<<1 #define rr o<<1|1 #define PI acos(-1.0) using namespace std; int dp[MAXN][MAXN], num[MAXN]; int main() { int n; while(Ri(n) != EOF) { for(int i = 0; i < n; i++) Ri(num[i]); CLR(dp, INF); for(int i = n-2; i >= 1; i--) { for(int j = i; j < n-1; j++) { if(i == j) dp[i][j] = num[i] * num[i-1] * num[i+1]; else { for(int k = i; k <= j; k++) { if(k == i) dp[i][j] = min(dp[i][j], dp[k+1][j] + num[k] * num[j+1] * num[i-1]); else if(k == j) dp[i][j] = min(dp[i][j], dp[i][k-1] + num[k] * num[j+1] * num[i-1]); else dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k+1][j] + num[k] * num[i-1] * num[j+1]); } } } } Pi(dp[1][n-2]); } return 0; }
相关文章推荐
- hdoj 2476 String painter 【区间dp】
- Codeforces 609B The Best Gift 【水题】
- Codeforces 607A:Chain Reaction 二分+递推
- 通过jquery实现页面的动画效果
- Codeforces 609A USB Flash Drives 【水题】
- 14443协议浅谈—TYPE_A与TYPE_B之比较[1]
- zepto
- ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 D. Delay Time
- HTML__post 和 get区别【URL】
- sublime text2 ctrl+b出现错误
- myeclipse 10和myeclipse2015软件和破解工具
- 蓝桥杯 算法提高 概率计算 (概率DP)
- 前端性能优化的14个规则,学会就偷着乐
- Oracle系列之权限
- 进击python第一篇:相遇
- OCI编程之个人浅见(二)
- ui-router系列文章
- Java的单例模式
- C#序列化匿名对象为XML
- Hdu 2039(水题) 解题报告