首页 > 代码库 > 题解:UESTC1217 The Battle of Chibi

题解:UESTC1217 The Battle of Chibi

2024-10-04 19:10:39   215人阅读

题意:对于一个N个数的序列求长度为M的上升子序列总数(结果取%)

         1<=N<=1000;1<=M<=N;1<=ai<=109.共1~100组数据，4000MS

解题思路:

考虑使用DP，先写出最朴素形式，DP[i][j][k]表示已判断到第i个数，序列长度为j，前一个数大小为k(已离散化)时方案数。复杂度为O(N*M*N)，直接上显然会T

优化方法较为明显，维护n个树状数组S[i]，S[j][k]表示到当前为止，最大数小于k且长度为j的序列总数，则每次对S[j](j从min(i,m)到1)进行操作：add(a[i]+1,tmp),其中tmp=ask(j-1,a[i]),则每次转移结束，S[j][k]保存从1到i的所有可能状态，复杂度O(NMlogN)

注意点:S数组转移时循环变量应采用倒序

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mo=1000000007;
const int maxn=1001;
int t,n,m,now,maxs,tmp,ans,gd,wz;
int s[2000][maxn+5];
int jl[2000],ys[2000];
int c[2000];

int lowbit(int x){return (x&(-x));}

void add(int u,int x,int value){
    for (int i=x;i<=maxs+1;i+=lowbit(i)){
        s[u][i]=(s[u][i]+value)%mo;
    }
}

int ask(int u,int x){
    {
        int sum=0;
        for (int i=x;i;i-=lowbit(i)) sum=(sum+s[u][i])%mo;
        return sum;
    }
}

bool cmd(int a,int b){
    return (jl[a]<jl[b]);
}

int main(){
    freopen("1.in","r",stdin);
    freopen("1.out1","w",stdout);
    scanf("%d",&t);
    for (int q=1;q<=t;++q){
        scanf("%d%d",&n,&m);maxs=0;
        for (int i=1;i<=n;++i) {scanf("%d",&jl[i]);ys[i]=jl[i];}
        for (int i=1;i<=n;++i) c[i]=i;
                sort(c+1,c+1+n,cmd);
           jl[c[1]]=1;wz=0;
           for (int i=2;i<=n;++i){
               if (ys[c[i]]==ys[c[i-1]]) ++wz;
               jl[c[i]]=i-wz;
        }
        maxs=n-wz;
        memset(s,0,sizeof(s));
        add(1,jl[1]+1,1);add(0,1,1);
        for (int i=2;i<=n;++i){
            for (int j=min(i,m);j>=1;--j) if (j>i) break; else {
              tmp=(ask(j-1,jl[i])) % mo;
              if (tmp!=0) add(j,jl[i]+1,tmp);
            }
        }
        //cout<<s[0][2]<<endl;
        int ans=ask(m,maxs+1) % mo;
        printf("Case #%d: %d\n",q,ans);
    }
    return 0;
}

 

题解:UESTC1217 The Battle of Chibi


联系
我们