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poj 1651 http://poj.org/problem?id=1651
http://poj.org/problem?id=1651Multiplication Puzzle
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6188 | Accepted: 3777 |
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.
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
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
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
610 1 50 50 20 5
Sample Output
3650
题意:
从中间选择一个数乘以左右俩边的数,第一个和最后一个不能选择,这个数就拿出来,怎样拿出来的顺序,使总和最小。
分析:abc sum=a*b*c;
abcd sum1=abc+acd;
sum2=bcd+abd;
sum=min(sum1,sum2)
dp[i][j]=min(dp[i][j],a[i]*a[g]*a[j]+dp[i][g]+dp[g][j]);(i<g<j)
也就是说igj,向该数列插入数字,是i和g之间或者是g和i之间。
#include<iostream>#include<cstring>#include<cstdio>using namespace std;int dp[105][105];int main(){ int i,j,k,n,a[105],g; while(~scanf("%d",&n)) { memset(dp,0,sizeof(dp)); for(i=0;i<n;i++) scanf("%d",&a[i]); for(k=2;k<n;k++)//保证最小有3位数。 { for(i=0;i<n-k;i++) { j=i+k; dp[i][j]=100000000;//求最小,初始化最大。 for(g=i+1;g<j;g++) dp[i][j]=min(dp[i][j],a[i]*a[g]*a[j]+dp[i][g]+dp[g][j]); } } printf("%d\n",dp[0][n-1]); } return 0;}/*610 1 50 50 20 53650*/
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