洛谷 P1155 双栈排序

题目传送门

解题思路:

首先考虑只有一个栈的时候如何解决这个问题。
就是对于一对位置
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<queue>
 5 #include<vector>
 6 #include<stack>
 7 #include<cstdlib>
 8 #define INF 0x7f7f7f7f
 9 
10 using namespace std;
11 
12 int n,a[1005],f[1005],color[1005];
13 vector<int> e[1005];
14 stack<int> s1,s2;
15 
16 inline void bfs(int s) {
17     queue<int> q;
18     q.push(s);
19     color[s] = 1;
20     while(!q.empty()) {
21         int now = q.front();
22         for(int i = 0;i < e[now].size(); i++) {
23             int next = e[now][i];
24             if(color[next] == -1) {
25                 color[next]= color[now] ^ 1;
26                 q.push(next); 
27             }
28             else if(color[next] != (color[now] ^ 1)) {
29                 printf("0");
30                 exit(0);
31                 return ; 
32             }
33         }
34         q.pop();
35     }
36 }
37 
38 int main()
39 { 
40     memset(color,-1,sizeof(color));
41     scanf("%d",&n);
42     for(int i = 1;i <= n; i++) 
43         scanf("%d",&a[i]);
44     f[n+1] = INF;
45     for(int i = n;i >= 1; i--) 
46         f[i] = min(f[i+1],a[i]);
47     for(int i = 1;i <= n; i++)
48         for(int j = i + 1;j <= n; j++)
49             if(a[i] > f[j+1] && a[i] < a[j])
50                 e[i].push_back(j),e[j].push_back(i);
51     for(int i = 1;i <= n; i++)
52         if(color[i] == -1)
53             bfs(i);
54     int cnt = 1;
55     for(int i = 1;i <= n; i++) {
56         if(color[i] == 1) {
57             s1.push(a[i]);
58             printf("a ");
59         }
60         else {
61             s2.push(a[i]);
62             printf("c ");
63         }
64         while((!s1.empty() && s1.top() == cnt) || (!s2.empty() && s2.top() == cnt)) {
65             if(!s1.empty() && s1.top() == cnt) {
66                 s1.pop();
67                 printf("b ");
68             }
69             else  {
70                 s2.pop();
71                 printf("d ");
72             }
73         ++cnt;
74         }
75     }
76     return 0;
77 }
//NOIP提高2008 T4
洛谷 P1155 双栈排序

题目传送门

解题思路:

AC代码:
