UVa #11925 Generating Permutations (习题8-7)

感觉把题目要求的过程反过来做会比较简单：把一个乱序的序列排成升序

做着做着觉得其实就是冒泡排序的改版嘛。如果最前面两个数不满足a<b，且 a != n-1 && b != 1(即：不是开头和结尾的交汇点)，则反转这两个数，然后挪动一位，再反复

复杂度大概是O(n^2)，还没有超过2*n^2的要求。这几天上mit的算法导论公开课，Eric说每次只能调换相邻两位元素的排序算法，复杂度下限是n^2。

Run Time: 0.192s

#define UVa  "8-7.11925.cpp"    //Generating Permutationschar fileIn[30] = UVa, fileOut[30] = UVa;#include<cstring>#include<cstdio>#include<vector>using namespace std;//Global Variables. Reset upon Each Case!const int maxn = 500;int n;int next[maxn], prev[maxn], a[maxn];vector<int> ans;/////int op(int code) {    ans.push_back(code);    if(code == 1) {        //swap first two.        int a = next[0], b = next[a];        int tmp = next[b];        next[a] = tmp;        prev[tmp] = a;        prev[a] = b;        next[b] = a;        prev[b] = 0;        next[0] = b;    }    else if(code == 2) {             //move to front.        int a = next[0], b = prev[n+1];        int tmp = next[a];        next[prev[b]] = n+1;        prev[n+1] = prev[b];        next[b] = a;        prev[b] = 0;        prev[a] = b;        next[0] = b;    }}void init() {    ans.clear();    memset(next, 0, sizeof(next));    memset(prev, 0, sizeof(prev));    for(int i = 1; i <= n; i ++) {        scanf("%d", &a[i]);        next[i] = i+1;        prev[i] = i-1;    }    next[n] = 0;    prev[0] = 0;    prev[n+1] = n;    next[0] = 1;}int invalid() {    int u = next[0], cnt = 0;    while(next[u] && next[u] != n + 1) {        if(a[u] && a[next[u]] && a[u] > a[next[u]]) return 1;        u = next[u];        cnt ++;    }    return 0;}int main() {    while(scanf("%d", &n) && n) {        init();        while(invalid()) {            int a1 = a[next[0]], a2 = a[next[next[0]]];            if(a1 > a2 && (a1 != n || a2 != 1)) op(1);            op(2);        }        for(int i = ans.size() - 1; i >= 0; i --) printf("%d", ans[i]);        printf("\n");    }    return 0;}