2018.2.20 寒假作業 A - Multiplication Puzzle
題目：
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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
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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
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Output must contain a single integer - the minimal score. |
sample
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6 10 1 50 50 20 5 |
Sample Output
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3650 |
題目大意：
一個整數序列包含N個1~100的整數（3<=N<=100），從中取出一個數並和相鄰兩邊的整數相乘，依次進行下去直到只剩下首尾兩個數為止，求最終的得到的和的最小值。兩邊的數不能取，且不重復選取。
思路：
dp[i][j]表示將第I個和第j個之間的數取完得到的和，那麽由k表示的數決定的子序列的寬度，依次遞增寬度，並且移動子序列，即改變i的值，然後對i到i+k之間的元素進行遍歷，比較dp[i][i+k]和先選取a[i]和a[j]之間的數，a[j]和a[i+k]之間的數，最後只剩下a[j]，這一過程中的得到的和dp[i][j]+dp[j][i+k]+a[i]*a[i+k]*a[j]與原來的和的大小。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include<cstdlib>
#include<ctype.h>
using namespace std;
typedef long long ll;
#define pi 3.14159265358979323846264338327
#define E 2.71828182846
#define INF 0x3f3f3f3f
#define maxn 555555
int a[110];
int dp[110][100];
int main() {
int T;
while (scanf("%d", &T) != EOF) {
int i, j, l;
for (i = 1; i <= T; i++) {
scanf("%d", &a[i]);
}
memset(dp, 0, sizeof(dp));
for (l = 2; l < T; l++) {
for (i = 2; i + l <= T + 1; i++) {
j = i + l - 1;
dp[i][j] = INF;
for (int k = i; k < j; k++) {
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + a[i - 1] * a[k] * a[j]);
}
}
}
printf("%d\n", dp[2][T]);
}
return 0;
}
2018.2.20 寒假作業 A - Multiplication Puzzle