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题目链接:
Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
/*4991 655MS 9664K 1701 B G++ 2014300227*/ #include <bits/stdc++.h> using namespace std; const int N=1e4+5; typedef long long ll; const ll mod=123456789; int n,m; ll sum[N],dp[102][N]; int lowbit(int x) { return x&(-x); } void update(int x,ll num) { while(x<=n) { sum[x]+=num; sum[x]%=mod; x+=lowbit(x); } } ll query(int x) { ll s=0; while(x>0) { s+=sum[x]; s%=mod; x-=lowbit(x); } return s; } struct node { int num,pos,c,d; }; node po[N]; int cmp1(node x,node y) { if(x.num==y.num)return x.pos<y.pos; return x.num<y.num; } int cmp2(node x,node y) { return x.pos<y.pos; } int main() { while(scanf("%d%d",&n,&m)!=EOF) { for(int i=1;i<=n;i++)scanf("%d",&po[i].num),po[i].pos=i; sort(po+1,po+n+1,cmp1); po[0].num=-1; for(int i=1;i<=n;i++) { if(po[i].num==po[i-1].num) { po[i].c=po[i-1].c; } else po[i].c=i;//po[i].c表示第一个跟po[i].num相同的数的位置; po[i].d=i;//表示po[i]插入时的位置; } sort(po+1,po+n+1,cmp2); for(int i=1;i<=n;i++) { dp[1][i]=1; update(po[i].d,1); } for(int i=2;i<=m;i++) { memset(sum,0,sizeof(sum)); for(int j=1;j<=n;j++) { if(po[j].c>1) dp[i][j]=query(po[j].c-1);//转移方程; else dp[i][j]=0; update(po[j].d,dp[i-1][j]);//把dp[i-1][j]更新上去; } } ll ans=0; for(int i=1;i<=n;i++) { ans+=dp[m][i]; ans%=mod; } printf("%lld\n",ans); } return 0; }
hdu-4991 Ordered Subsequence(dp+树状数组)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5399769.html