Time limit1000 ms Memory limit65536 kB The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the ...
Time limit1000 ms
Memory limit65536 kB
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
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
6 10 1 50 50 20 5
Sample Output
3650
題意：頭尾不動，每次都取一個數，使得它和它的兩邊的數相乘為去掉這個數的價值，問直到只剩下頭尾所需的最小價值
題解：求出每個區間的最小值，直到擴充到整個區間，區間dp
#include<iostream> #include<algorithm> #include<cstring> #include<sstream> #include<cmath> #include<cstdlib> #include<queue> #include<stack> using namespace std; #define PI 3.14159265358979323846264338327950 #define INF 0x3f3f3f3f; int n,i,j,k,l; int dp[110][110]; int a[1010]; void solve() { for(l=2;l<=n;l++) { for(i=2;i+l<=n+1;i++) { j=i+l-1; dp[i][j]=INF; for(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][n]); } int main() { while(~scanf("%d",&n)) { memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) cin>>a[i]; solve(); } }
