POJ 3617 Best Cow Line (贪心)

Posted 2020-08-05 dwtfukgv

题意：给定一行字符串，让你把它变成字典序最短，方法只有两种，要么从头部拿一个字符，要么从尾部拿一个。

析：贪心，从两边拿时，哪个小先拿哪个，如果一样，接着往下比较，要么比到字符不一样，要么比完，也就是说从头部和尾部拿都一样，那么就随便拿一个了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <list>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
    while(scanf("%d", &n) == 1){
        string s;  char ss[5];
        for(int i = 0; i < n; ++i){
            scanf("%s", ss);
            s.push_back(ss[0]);
        }
        string ans;
        for(int i = 0, j = n-1; i <= j; ){
            if(s[i] > s[j]) ans.push_back(s[j]), --j;
            else if(s[i] < s[j])  ans.push_back(s[i]), ++i;
            else{
                bool ok = false;
                for(int k = i, l = j; k <= l; ){
                    if(k == l)  ans.push_back(s[i]), ++i, ++k, ok = true;
                    else if(s[k] == s[l])  ++k, --l;
                    else{
                        if(s[k] < s[l])  ans.push_back(s[i]), ++i;
                        else  ans.push_back(s[j]), --j;
                        ok = true;
                        break;
                    }
                }
                if(!ok) ans.push_back(s[i]), ++i;
            }
        }
        for(int i = 0; i < ans.size(); ++i){
            if(i && i % 80 == 0) printf("\n");
            printf("%c", ans[i]);
        }
        printf("\n");
    }
    return 0;
}