hdoj 3183 A Magic Lamp 【RMQ区间取数】

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2525    Accepted Submission(s): 989

Problem Description

Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question, and then the genie will realize one of her dreams. 
The question is: give you an integer, you are allowed to delete exactly m digits. The left digits will form a new integer. You should make it minimum.
You are not allowed to change the order of the digits. Now can you help Kiki to realize her dream?

 

Input

There are several test cases.
Each test case will contain an integer you are given (which may at most contains 1000 digits.) and the integer m (if the integer contains n digits, m will not bigger then n). The given integer will not contain leading zero.

 

Output

For each case, output the minimum result you can get in one line.
If the result contains leading zero, ignore it. 

 

Sample Input

178543 4 1000001 1100001 212345 254321 2

 

Sample Output

1310123321

 

题意：给定一个用字符串表示的大数str和一个数M，让你从左到右取出|str|-M个数，使得取出的数最小，并输出。【去掉前导0】

思路：RMQ区间取数，维护的是区间下标最小的最小值。

dp[i][j]记录以i开头的长度为1<<j的区间里面下标最小的最小值。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-4#define MAXN (1000+10)#define MAXM (1000000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 100000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;int dp[MAXN][32];int a[MAXN];int N;void RMQ_init(){    for(int i = 1; i <= N; i++)        dp[i][0] = i;    for(int j = 1; (1<<j) <= N; j++)        for(int i = 1; i + (1<<j)-1 <= N; i++)            dp[i][j] = a[dp[i][j-1]] <= a[dp[i+(1<<(j-1))][j-1]] ? dp[i][j-1] : dp[i+(1<<(j-1))][j-1];}int Query(int L, int R){    int k = 0;    while((1<<(k+1)) <= R-L+1) k++;    return a[dp[L][k]] <= a[dp[R-(1<<k)+1][k]] ? dp[L][k] : dp[R-(1<<k)+1][k];}int main(){    char str[1010]; int M;    while(scanf("%s%d", str, &M) != EOF)    {        N = strlen(str);        for(int i = 0; i < N; i++)            a[i+1] = str[i] - '0';        RMQ_init();        int top = 0, pos = 1;        for(int i = M+1; i <= N; i++)        {            pos = Query(pos, i);            str[top++] = a[pos] + '0';            pos++;        }        pos = -1;        for(int i = 0; i < top; i++)            if(str[i] != '0')            {                pos = i;                break;            }        if(pos == -1)            printf("0\n");        else        {            for(; pos < top; pos++)                printf("%c", str[pos]);            printf("\n");        }    }    return 0;}