POJ 1651 Multiplication Puzzle【区间dp】

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Multiplication Puzzle

POJ - 1651

思路：求出每个区间的最小值，一直扩散到整个区间

设dp[l][r]表示区间[l,r]的最优解，则状态转移如下：

1、当r-l=2时，dp[l][r] = num[l]*num[l+1]*num[r]；

2、当r-l>2时，对区间的最后一个被拿走的数进行枚举，则dp[l][r] = min(dp[l][r], dp[l][i]+dp[i][r]+num[l]*num[i]*num[r])

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))const int M=1e10+10;const int inf=0x3f3f3f3f;const int mod =1e9+7;int n,a[105],dp[105][105];int i,j,k,l;int main(){    while(~scanf("%d",&n))  {        for(i = 1; i<=n; i++)            scanf("%d",&a[i]);        ms(dp,0);        for(l = 2; l<n; l++) {            for(i = 2; i+l<=n+1; i++)  {                j = i+l-1;                dp[i][j] = M;                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]);    }    return 0;}

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