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：操做字符串，通過調整字符串的順序獲得一個新的字符串，新的字符串的數值是比原字符串的數大的最小值;

題解：因爲是最小值，咱們能夠從後往前遍歷字符串，當str[i]>str[i-1]時,pos=i-1，此位置就是咱們要調整字符串的起始位置(說明咱們能夠調整順序使原來的字符串的值增大），而後從咱們要截取的字串中找到第一個比上pos大的值交換，而後將字串排序便可，若沒有條件知足str[i]>str[i-1],則經過調整字符順序沒有比原串更大的數值。

#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[1001];
int dis[500][500];
int di[4][2]= {{-1,0},{1,0},{0,-1},{0,1}};
map<ll,ll>::iterator it;
int main()
{
    cin>>m;
    while(m--)
    {
        string str;
        cin>>n>>str;
        int pos=-1;
        for(int i=str.size()-1; i>=1; i--)
        {
            if(str[i]>str[i-1])
            {
                pos=i-1;
                break;
            }
        }
        if(pos==-1)
        {
            cout<<n<<" "<<"BIGGEST"<<endl;
            continue;
        }
        string s,arr;
        s=str.substr(pos);
        int cnt=s[0];
        for(int i='0'; i<='9'; i++)
        {

            if(s.find(i)!=string::npos&&i>cnt)
            {
                swap(s[0],s[s.find(i)]);
                break;
            }
        }
        sort(s.begin()+1,s.end());
        cout<<n<<" "<<str.substr(0,pos)<<s<<endl;;
    }

}

View Code