UVALIVE 4556 The Next Permutation

4556 The Next Permutation
For this problem, you will write a program that takes a (possibly long) string of decimal digits, and
outputs the permutation of those decimal digits that has the next larger value (as a decimal number)
than the input number. For example:
123 -> 132
279134399742 -> 279134423799
It is possible that no permutation of the input digits has a larger value. For example, 987.
Input
The first line of input contains a single integer P , (1 ≤ P ≤ 1000), which is the number of data sets that
follow. Each data set is a single line that contains the data set number, followed by a space, followed
by up to 80 decimal digits which is the input value.
Output
For each data set there is one line of output. If there is no larger permutation of the input digits,
the output should be the data set number followed by a single space, followed by the string ‘BIGGEST’.
If there is a solution, the output should be the data set number, a single space and the next larger
permutation of the input digits.
Sample Input
3
1 123
2 279134399742
3 987
Sample Output
1 132
2 279134423799
3 BIGGEST

題意：操作字符串，通過調整字符串的順序得到一個新的字符串，新的字符串的数值是比原字符串的数大的最小值;

题解：由于是最小值，我们可以从后往前遍历字符串，当str[i]>str[i-1]时,pos=i-1，此位置就是我们要调整字符串的起始位置(说明我们可以调整顺序使原来的字符串的值增大），然后从我们要截取的字串中找到第一个比上pos大的值交换，然后将字串排序即可，若没有条件满足str[i]>str[i-1],则通过调整字符顺序没有比原串更大的数值。

#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#include<iostream>
#include<algorithm>
#include<map>
#include<vector>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
int m,n,k,p;
int ans[];
int dis[][];
int di[][]= {{-,},{,},{,-},{,}};
map<ll,ll>::iterator it;
int main()
{
    cin>>m;
    while(m--)
    {
        string str;
        cin>>n>>str;
        int pos=-;
        for(int i=str.size()-; i>=; i--)
        {
            if(str[i]>str[i-])
            {
                pos=i-;
                break;
            }
        }
        if(pos==-)
        {
            cout<<n<<" "<<"BIGGEST"<<endl;
            continue;
        }
        string s,arr;
        s=str.substr(pos);
        int cnt=s[];
        for(int i=''; i<=''; i++)
        {

            if(s.find(i)!=string::npos&&i>cnt)
            {
                swap(s[],s[s.find(i)]);
                break;
            }
        }
        sort(s.begin()+,s.end());
        cout<<n<<" "<<str.substr(,pos)<<s<<endl;;
    }

}