hdu3506Monkey Party
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链接:http://acm.hdu.edu.cn/showproblem.php?pid=3506
题意:给定n个猴子坐成环,每个猴子介绍自己需要花费a[i]分钟。每次可以选择两个相邻的两个猴子,然后让他们两个的朋友互相认识花费的时间是这两堆猴子介绍自己的总和,其实就是环形石子合并啦。
分析:很明显我们是可以区间dp的,设dp[i][j]表示i~j个猴子全部互相认识需要花费的最少时间,那么dp[i][j]=min(dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]),i<=k<j。这样是O(n^3)的,我们用四边形优化一下就能到O(n^2)啦,因为sum[j]-sum[i-1]显然是满足四边形优化的。四边形优化不会的转poj1160Post Office
代码:
#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=1010;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000007;const int INF=1000000010;const ll MAX=1000000000000;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;ll a[2*N],sum[2*N],s[2*N][N],dp[2*N][N];int main(){ int i,j,k,n; ll ans; while (scanf("%d", &n)!=EOF) { for (i=1;i<=n;i++) { scanf("%I64d", &a[i]);a[i+n]=a[i]; } for (i=1;i<=2*n;i++) sum[i]=sum[i-1]+a[i]; memset(dp,0x7f,sizeof(dp)); for (i=1;i<=2*n;i++) dp[i][i]=0,s[i][i]=i; for (i=1;i<n;i++) for (j=2*n-i;j;j--)//[j,j+i] for (k=s[j][j+i-1];k<=s[j+1][j+i];k++) if (dp[j][k]+dp[k+1][j+i]+sum[j+i]-sum[j-1]<=dp[j][j+i]) { dp[j][j+i]=dp[j][k]+dp[k+1][j+i]+sum[j+i]-sum[j-1];s[j][j+i]=k; } ans=MAX; for (i=1;i<=n;i++) ans=min(ans,dp[i][i+n-1]); printf("%I64d\n", ans); } return 0;}
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