51Nod 1022（四边形优化dp）

来源：互联网发布：腾讯游戏平台mac版本编辑：程序博客网时间：2024/05/19 07:10

传送门
题解：由数据范围知复杂度大约为O(n^2)，所以考虑四边形优化dp，尽快补上四边形优化的讲解。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN=2004,INF=0x3f3f3f3f;int n,a[MAXN],dp[MAXN][MAXN],s[MAXN][MAXN],sum[MAXN][MAXN];int main() {    while (~scanf("%d",&n)) {        memset(sum,0,sizeof(sum));        for (int i=1;i<=n;++i) scanf("%d",&a[i]),a[n+i]=a[i];        for (int i=1;i<=(n<<1);++i) dp[i][i]=0,s[i][i]=i;        for (int i=1;i<=(n<<1);++i)            for (int j=i;j<=min(i+n,n<<1);++j)                sum[i][j]=sum[i][j-1]+a[j];        for (int len=2;len<=n;++len)            for (int i=1;i+len-1<=(n<<1);++i) {                int j=i+len-1;                dp[i][j]=INF;                for (int k=s[i][j-1];k<=s[i+1][j];++k) {                    if (dp[i][j]>dp[i][k]+dp[k+1][j]+sum[i][j])                        dp[i][j]=dp[i][k]+dp[k+1][j]+sum[i][j],s[i][j]=k;                }            }        int ans=INF;        for (int i=1;i<=n;++i)            ans=min(ans,dp[i][i+n-1]);        printf("%d\n",ans);    }    return 0;}

阅读全文

0 0