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  • BZOJ1049: [HAOI2006]数字序列

    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1049

    题解:

    ydc的题解:http://pan.baidu.com/share/link?uk=2651016602&shareid=1490516411

    第二问比较神?考场上敢去猜结论?

    代码:

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<cmath>
     4 #include<cstring>
     5 #include<algorithm>
     6 #include<iostream>
     7 #include<vector>
     8 #include<map>
     9 #include<set>
    10 #include<queue>
    11 #include<string>
    12 #define inf 1000000000
    13 #define maxn 50000+5
    14 #define maxm 100000+5
    15 #define eps 1e-10
    16 #define ll long long
    17 #define pa pair<int,int>
    18 #define for0(i,n) for(int i=0;i<=(n);i++)
    19 #define for1(i,n) for(int i=1;i<=(n);i++)
    20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
    21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
    22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
    23 #define mod 1000000007
    24 using namespace std;
    25 inline int read()
    26 {
    27     int x=0,f=1;char ch=getchar();
    28     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    29     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
    30     return x*f;
    31 }
    32 int n,a[maxn],b[maxn],f[maxn],next[maxn],last[maxn];
    33 ll g[maxn],t[maxn];
    34 int main()
    35 {
    36     freopen("input.txt","r",stdin);
    37     freopen("output.txt","w",stdout);
    38     n=read();
    39     for1(i,n+2)b[i]=inf;
    40     for2(i,2,n+1)a[i]=read()-i;
    41     a[1]=-inf;a[n+2]=inf-1;
    42     n+=2;
    43     for1(i,n)
    44     {
    45         f[i]=upper_bound(b+1,b+n+1,a[i])-b;
    46         b[f[i]]=a[i];
    47     }
    48     cout<<n-f[n]<<endl;
    49     for3(i,n,1)next[i]=last[f[i]],last[f[i]]=i;
    50     memset(g,127,sizeof(g));
    51     for1(i,n)
    52     {
    53         if(f[i]==1){g[i]=0;continue;}
    54         t[i]=0;
    55         for3(j,i-1,last[f[i]-1])t[j]=t[j+1]+abs(a[j]-a[i]);
    56         for(int j=last[f[i]-1];j;j=next[j])if(a[j]<=a[i])
    57         {
    58             if(j+1==i){g[i]=min(g[i],g[j]);continue;}
    59             ll tot=0;
    60             for2(k,j,i-1)
    61             {
    62                 tot+=abs(a[j]-a[k]);
    63                 g[i]=min(g[i],g[j]+tot+t[k+1]);
    64             }
    65         }
    66     }
    67     cout<<g[n]<<endl;
    68     return 0;
    69 }
    View Code
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  • 原文地址:https://www.cnblogs.com/zyfzyf/p/4246861.html
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