[POJ 3617]Best Cow Line

Description

FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual”Farmer of the Year” competition. In this contest every farmer arranges his cows in a line and herds them past the judges.

The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows’ names.

FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.

FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he’s finished, FJ takes his cows for registration in this new order.

Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.

Input

  • Line 1: A single integer: N
  • Lines 2~N+1: Line i+1 contains a single initial (‘A’..’Z’) of the cow in the ith position in the original line

Output

The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows (‘A’..’Z’) in the new line.

Sample Input

6
A
C
D
B
C
B

Sample Output

ABCBCD

Hint

使用贪心算法,不断将较小的字符放入新字符串的末尾,这样我们就能构造一个字典序最小的串。对于首尾字符相同的情景,取哪个都可以。
注意题目要求,每 80 字符换行。

题解

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <cstring>
//#include <regex>
#define online
#define INF 2010
using namespace std;

int n, head, tail;
char str[INF];
string str2;
int chk_head,chk_tail;
inline bool check()
{
    #ifdef offline
    printf("new check\n");
    #endif
    chk_head = head;
    chk_tail = tail;
    while (chk_head <= chk_tail)
    {
        if (str[chk_head] == str[chk_tail])
        {
            #ifdef offline
            printf("str[chk_head] %d %c== str[chk_tail] %d %c\n", chk_head, str[chk_head], chk_tail, str[chk_tail]);
            #endif
            chk_head++;
            chk_tail--;
        }
        else
        {
            if (str[chk_head] > str[chk_tail])
            {
                str2 = str2 + str[tail];
                tail--;
                #ifdef offline
                printf("check end at tail\n");
                #endif
                return 1;//tail
            }
            else
            {
                str2 = str2 + str[head];
                head++;
                #ifdef offline
                printf("check end at head\n");
                #endif
                return 0;//head
            }
        }
    }
    str2 = str2 + str[tail];
    tail--;
    #ifdef offline
    printf("check end at outside\n");
    #endif
}
inline bool solve()
{
    #ifdef offline
    printf("new solve\n");
    #endif
    if (str[head] > str[tail]){
        str2 = str2 + str[tail];
        tail--;
        #ifdef offline
        printf("solve 1\n");
        #endif
    }
    else
    {
        #ifdef offline
        printf("solve 0\n");
        #endif

        if (str[head] == str[tail] && head != tail){
            #ifdef offline
            printf("head == tail ==  %d\n", head);
            #endif
            check();
        }
        else
        {
            str2 = str2 + str[head];
            head++;
        }
    }
    if (head > tail)
        return 0;
    else
        solve();
}

int main()
{
    //srand((unsigned int)(time(NULL)));
    //freopen("","r",stdin);
    //freopen("","W",stdout);
    scanf("%d", &n);
    head = 0;
    tail = n-1;
    for (int i = 0; i < n; i++)
        scanf("%c%c", &str[i], &str[i]);
    solve();
    //cout<<str2;//poisonous output format
    for (int i = 0; i < str2.length(); i++)
    {
        if ((i + 1) % 80 == 0){
            if (i + 1 == n)
                printf("%c", str2[i]);
            else
                printf("%c\n", str2[i]);
        }
        else
            printf("%c", str2[i]);
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

[POJ 2386]Lake Counting

Description

有一个大小为 N × M 的园子,雨后积起了水。八连通的积水被认为是连接在一起的。请求出
园子里总共有多少水洼?(八连通指的是下图中相对 W 的*的部分)

* * *
* W *
* * *

Input

第 1 行 两个数 N, M
第2~n行 各 M 个数

Sample Input

10 12
W........WW. 
.WWW.....WWW 
....WW...WW. 
.........WW. 
.........W.. 
..W......W.. 
.W.W.....WW. 
W.W.W.....W. 
.W.W......W. 
..W.......W.

Sample Output

3

题解

通过二维布尔表来记录积水,搜索邻接坐标的状态并重置以避免重复计数。

#include "cstdio"
#include "cmath"
#include "cstdlib"
#include "iostream"
#include "cstring"
#define INF 110
#define online
using namespace std;
int count = 0;
int n,m;
bool map[INF][INF];
inline bool handle(int n1,int m1,bool sign)
{
    //endowed W as 1, NULL and * as 0
    //use sign to distinguish the original requestion and derived requestion
    //True->Origin
    if (n1 < 0 || m1 < 0 || n1 >= n || m1 >= m)
    {
        #ifdef debug
        printf("n1 %d m1 %d n %d m %d out\n", n1, m1, n, m);
        #endif
        return 0;
    }
    if (map[n1][m1])
    {
        #ifdef debug
        printf("matched\n");
        #endif
        map[n1][m1] = 0;
        if (sign)
            count++;
        handle(n1,m1-1,0);
        handle(n1,m1+1,0);
        handle(n1+1,m1,0);
        handle(n1-1,m1,0);
        handle(n1-1,m1-1,0);
        handle(n1+1,m1+1,0);
        handle(n1+1,m1-1,0);
        handle(n1-1,m1+1,0);
    }

}

int main()
{
    //initialize
    scanf("%d%d", &n, &m);
    char c, str[m+1];
    memset(map,0,sizeof(map));
    //endowed * as 0, W as 1
    for (int i = 0; i < n; i++)
    {
        scanf("%s", &str);
        for (int j = 0; j < m; j++)
            map[i][j] = str[j] == '.' ? 0 : 1;

    }

    #ifdef debug
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++)
            printf("%d", map[i][j]);
        printf("\n");
    }
    #endif

    //handle
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            handle(i,j,1);
    printf("%d\n", count);

    return 0;
}