给1个排列 求下1个排列 按字典序
跟普通排列不同的地方就是 有相同的数字
那末就把普通的1改就完事
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#include <map>
#include <ctime>
#define MAXN 222
#define MAXM 6122222
#define INF 1000000001
using namespace std;
int n, c[1111];
char s[1111];
void get_next() {
int pos1 = n – 1;
while(pos1 > 0 && c[pos1] >= c[pos1 + 1]) pos1–; //>变成了>=
int pos2 = n;
while(c[pos2] <= c[pos1]) pos2–; //<变成了<=
swap(c[pos2], c[pos1]);
int l = pos1 + 1, r = n;
while(l < r) {
swap(c[l], c[r]);
l++; r–;
}
}
int main() {
int T, cas;
scanf("%d", &T);
while(T–) {
scanf("%d", &cas);
scanf("%s", s);
n = strlen(s);
for(int i = 0; i < n; i++) {
c[i + 1] = s[i] – '0';
}
printf("%d ", cas);
bool flag = 1;
for(int i = 1; i < n; i++)
if(c[i] < c[i + 1]) flag = 0;
if(flag) {
printf("BIGGEST
");
continue;
}
get_next();
for(int i = 1; i <= n; i++) printf("%d", c[i]);
printf("
");
}
return 0;
}
波比源码 » POJ 3785 The Next Permutation 全排列字典序法
