bzoj4606 DNA

题意有点复杂，就自己去看吧。

题解   

f[i][j][k]表示第i为，为-j范式，当前位为k的方案数，从后往前dp一遍。

然后再贪心的从前往后填数。但注意当前填的数已经是-cur范式，计算方案的时候要用k-cur来算。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100020

typedef long long ll;
ll f[maxn][11][4],R;
int mp[200],a[maxn],ans[maxn];
int n,k;
char ch[maxn],re[5];

void init(){
    if ( a[n] == 5 ){
        for (int l = 0 ; l <= 3 ; l++){
            for (int i = 1 ; i <= k ; i++){
                f[n][i][l] = 1;
            }
        }
    }
    else{
        for (int i = 1 ; i <= k ; i++)f[n][i][a[n]] = 1; 
    }
    for (int i = n - 1 ; i ; i--){
        if ( a[i] < 5 ){
            for (int j = 1 ; j <= k ; j++){
                for (int l = 0 ; l <= 3 ; l++){
                    if ( a[i] >= l ) f[i][j][a[i]] += f[i + 1][j][l]; //如果有这种转移就不能在用f[i + 1][j - 1][l]在更新了。因为f[i + 1][j][l]已经包含了它的方案数，否则会算重
                    else f[i][j][a[i]] += f[i + 1][j - 1][l];
                }
            }
        }
        else{
            for (int l = 0 ; l <= 3 ; l++){
                for (int j = 1 ; j <= k ; j++){
                    for (int p = 0 ; p <= 3 ; p++){
                        if ( l >= p ) f[i][j][l] += f[i + 1][j][p];
                        else f[i][j][l] += f[i + 1][j - 1][p];
                    }
                }
            }
        }
    }
/*    for (int i = 1 ; i <= n ; i++)
       for (int l = 0 ; l <= 3 ; l++)    
        for (int j = 1 ; j <= k ; j++){
            f[i][j][l] += f[i][j - 1][l];
        }*/
}
void solve(){
    int cur = 1;
    ans[0] = 5;
    for (int i = 1 ; i <= n ; i++){
        if ( a[i] == 5 ){
            ll sum[4];
            for (int l = 0 ; l <= 3 ; l++){
                if ( ans[i - 1] < l ) sum[l] = f[i][k - cur][l];
                else sum[l] = f[i][k - cur + 1][l];    
            }
            for (int l = 3 ; l >= 0 && R > 0 ; l--){
                ans[i] = l , R -= sum[l];
            }
            R += sum[ans[i]];
            if ( ans[i - 1] < ans[i] ) cur++;
        }    
        else{
            ans[i] = a[i];
            if ( ans[i - 1] < ans[i] ) cur++;
        }
    }
    for (int i = 1 ; i <= n ; i++) printf("%c",re[ans[i]]);
    printf("
");
}
int main(){
    freopen("input.txt","r",stdin);
    mp['A'] = 3 , mp['C'] = 2 , mp['G'] = 1 , mp['T'] = 0 , mp['N'] = 5;
    re[3] = 'A' , re[2] = 'C' , re[1] = 'G' , re[0] = 'T';
    scanf("%d %d %lld",&n,&k,&R);
    scanf("%s",ch + 1);
    for (int i = 1 ; i <= n ; i++) a[i] = mp[(int)ch[i]];
    init();
    solve();
}