POJ 1651 Multiplication Puzzle 区间DP

Multiplication Puzzle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10664 Accepted: 6659

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

610 1 50 50 20 5

Sample Output

3650

Source

Northeastern Europe 2001, Far-Eastern Subregion

给n个数字，要删掉中间n-2个，每次删除的代价是自己的数乘左右两边的数的乘积，问最小代价是多少。

显然如果删去的区间不连续，则代价是彼此独立的，所以适合用区间DP解决。

dp[i][j]表示删去 i 到 j 连续的一段区间的最小代价，设最后删除的数字的位置为k，则

dp[j][j+i]=min(dp[j][k-1]+dp[k+1][j+i]+a[k]*a[j-1]*a[j+i+1])

除了这种情况，还需要考虑k正好在区间两边的情况。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int maxn=105,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const ld pi=acos(-1.0L);ll dp[maxn][maxn],a[maxn];int main() {int n,i,j,k;scanf("%d",&n);for (i=1;i<=n;i++) scanf("%lld",&a[i]);meminf(dp);for (i=2;i<n;i++) dp[i][i]=a[i-1]*a[i]*a[i+1]; for (i=1;i<=n-2;i++) {for (j=2;j+i<n;j++) {dp[j][j+i]=min(dp[j][j+i],dp[j+1][j+i]+a[j]*a[j-1]*a[j+i+1]);dp[j][j+i]=min(dp[j][j+i],dp[j][j+i-1]+a[j+i]*a[j-1]*a[j+i+1]);for (k=j+1;k<j+i;k++) {dp[j][j+i]=min(dp[j][j+i],dp[j][k-1]+dp[k+1][j+i]+a[k]*a[j-1]*a[j+i+1]);}//cout << j << ' ' << j+i << ' ' << dp[j][j+i] << endl;}}printf("%lld\n",dp[2][n-1]);return 0;}