• 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
    ",ans);
    }
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  • 原文地址:https://www.cnblogs.com/Profish/p/11200467.html
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