dp四边形优化

如果dp[i][j]=min(dp[i][k]+dp[k+1][j]+w[i][j]);且满足dp[a][c]+dp[b][d]<=dp[a][d]+dp[c][d](a<b<=c<d);

那么dp具有四边形不等式性质

另外如果可以证明w[i][j]满足单调性和四边形不等式性质,那么dp也具有四边形不等式性质

单调性:w[i][j]<=w[i][j+1]<=w[i+1][j+1]

http://www.51nod.com/Challenge/Problem.html#problemId=1022

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define LNF 0x3f3f3f3f3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=1e6+8;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+7;
const ULL base=1e7+7;
using namespace std;
LL dp[2008][2008];
LL s[2008][2008];
LL a[2008];
int main(){
    int n;
    int x;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        a[i+n]=a[i];
    }
    n*=2;
    for(int i=1;i<=n;i++){
        a[i]+=a[i-1];
    }
    memset(dp,LNF,sizeof(dp));
    for(int i=1;i<=n;i++){
        s[i][i]=i;
        dp[i][i]=0;
    }
    for(int i=n;i>=1;i--){
        for(int j=i+1;j<=n;j++){
            for(int k=s[i][j-1];k<=s[i+1][j];k++){
                if(dp[i][j]>dp[i][k]+dp[k+1][j]+a[j]-a[i-1]){
                    dp[i][j]=dp[i][k]+dp[k+1][j]+a[j]-a[i-1];
                    s[i][j]=k;
                }
            }
        }
    }
    n/=2;
    LL ans=LNF;
    for(int i=1;i<=n;i++){
        ans=min(ans,dp[i][i+n-1]);
    }
    printf("%lld\n",ans);
}

猜你喜欢

转载自www.cnblogs.com/Profish/p/11200467.html