3072: [Pa2012]Two Cakes

来源：互联网发布：linux性能分析工具编辑：程序博客网时间：2024/06/06 18:57

有dp[i][j]=min(d[i-1][j],d[i][j-1])+1 –> a[i]==b[j]
dp[i][j]=d[i-t+1][j-t+1]+t —> t为i,j向前最长的不同序列；
但真正需要记录的状态为O(n)；

#include<bits/stdc++.h>#define rep(i,k,n) for(int i=k;i<=n;i++)using namespace std;const int N=1e6+7;void sc(int& x){    char c=getchar();x=0;    for(;c>'9'||c<'0';c=getchar());    for(;c<='9' && c>='0';c=getchar())x=10*x+c-'0';}vector<int> dif[N<<1];int n,a[N],b[N],pos[N],dp[N];inline int find(int id, int x) {    int l = 0, r = dif[id].size() - 1, res = -1;    while(l <= r) {        int mid = l + r >> 1;        if(dif[id][mid] <= x) res = mid, l = mid + 1;        else r = mid - 1;    }    return ~res ? dif[id][res] : 0;}inline int Dp(int x,int y){    if(!x||!y)return x+y;    if(a[x]==b[y])return dp[x] ? dp[x] : (dp[x]=min(Dp(x-1,y),Dp(x,y-1))+1);    int t=find(x-y+n,x);    return t ? Dp(t,y-x+t)+x-t : max(x,y);}int main(){    sc(n);rep(i,1,n)sc(a[i]);rep(i,1,n)sc(b[i]),pos[b[i]]=i;    rep(i,1,n)dif[i-pos[a[i]]+n].push_back(i);    printf("%d\n",Dp(n,n));}

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