标签：put scan namespace enter ace string lin algo algorithm
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
Input
Output
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
1.和石子归并差不多的问题。
正确的代码：
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 using namespace std; 6 typedef long long ll; 7 8 int N; 9 int dp[105][105],a[105]; 10 11 int main() 12 { scanf("%d",&N); 13 for(int i=1;i<=N;i++) scanf("%d",&a[i]); 14 memset(dp,0,sizeof(dp)); 15 for(int i=1;i<N-1;i++) dp[i][i+2]=a[i]*a[i+1]*a[i+2]; 16 for(int len=3;len<N;len++){ 17 for(int i=1;i<=N&&i+len<=N;i++){ 18 int j=len+i; 19 for(int k=i+1;k<j;k++){ 20 if(dp[i][j]==0) dp[i][j]=dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]; 21 dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]); 22 } 23 } 24 } 25 printf("%d\n",dp[1][N]); 26 }
错误的代码：调整k的时候遗漏某些情况，如1 2 3 4 5，取2之后再取4，这种情况会被漏掉！
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 using namespace std; 6 typedef long long ll; 7 8 const int INF=1000000000; 9 10 int N; 11 int dp[105][105],a[105]; 12 13 int main() 14 { scanf("%d",&N); 15 for(int i=1;i<=N;i++) scanf("%d",&a[i]); 16 memset(dp,0,sizeof(dp)); 17 for(int i=1;i<N-1;i++) dp[i][i+2]=a[i]*a[i+1]*a[i+2]; 18 for(int len=3;len<N;len++){ 19 for(int i=1;i<=N&&i+len<=N;i++){ 20 int j=len+i; 21 dp[i][j]=INF; 22 for(int k=i;k<j;k+=len-1){ 23 int tem; 24 if(k==i) tem=dp[i][k]+dp[k+1][j]+a[i]*a[k+1]*a[j]; 25 else tem=dp[i][k]+dp[k+1][j]+a[i]*a[k]*a[j]; 26 if(dp[i][j]>tem) dp[i][j]=tem; 27 } 28 } 29 } 30 printf("%d\n",dp[1][N]); 31 }
Multiplication Puzzle POJ - 1651
标签：put scan namespace enter ace string lin algo algorithm
原文地址：http://www.cnblogs.com/zgglj-com/p/6914838.html
