HDU 4352 XHXJ's LIS 数位dp lis

---

[TOC]

#题目链接
HDU 4352 XHXJ's LIS
#题解
对于lis求的过程
对一个数列，都可以用nlogn的方法来的到它的一个可行lis
对这个logn的方法求解lis时用的数组进行装压
预处理的到这个的转移
数位dp转移的时候直接得到下一位的lis状态
#代码

#include<set> 
#include<cstdio> 
#include<cstring> 
#include<algorithm> 
#define gc getchar()
#define pc putchar 
#define LL long long 
inline LL read() { 
	LL x = 0,f = 1; 
	char c = gc; 
	while(c < '0' || c > '9') c = gc; 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc; 
	return x * f;  
} 
void print(LL x) { 
	if(x < 0) { 
		pc('-'); 
		x = -x; 
	} 
	if(x >= 10) print(x / 10); 
	pc(x % 10 + '0'); 
} 
const int maxn = (1 << 10) + 7; 
int cnt[maxn]; 
int nxt[maxn][21]; //状态s插入j之后的最优可行方案 
int get(int x,int y) { 
	for(int i = y;i < 10;++ i) { 
		if(x & (1 << i)) { 
			return x ^ (1 << i) | (1 << y); 
		} 
	} 
	return x | (1 << y); 
} 
void pre() { 
	for(int i = 0;i < (1 << 10); ++ i) { 
		for(int j = 0;j < 10;++ j) { 
			if(i & (1 << j) ) cnt[i] ++; 
			nxt[i][j] = get(i,j); 
		} 
	} 
} 
LL dp[21][maxn][21]; 
int limit[22]; 
LL dfs(int k,int len,int num,bool flag,bool zero) {
	if(len < 0) 
		return cnt[num] == k; 
	if(!flag && dp[len][num][k] != -1) return dp[len][num][k]; 
	LL ret = 0; 
	int lim = flag ? limit[len] : 9; 
	for(int i = 0;i <= lim;++ i) 
		ret += dfs(k,len - 1,(zero && i == 0) ? num : nxt[num][i],(flag && i == lim),(zero && i == 0)); 
	if(!flag) 
	dp[len][num][k] = ret; 
	return ret; 
} 
LL solve(LL n,int k) { 
	int pos = 0; 
	while(n) { 
		limit[pos ++] = n % 10; 
		n /= 10; 
	} 
	return dfs(k,pos - 1,0,1,1); 
} 
int main() { 
	memset(dp,-1,sizeof dp); 
	int T = read(); 
	pre(); 	
	for(int i = 1;i <= T;++ i) { 
		LL L = read(),R = read(),k = read(); 
		printf("Case #%d: ",i); 
		print(solve(R,k) - solve(L - 1,k)); 
		pc('\n'); 
	} 
	return 0; 
}