POJ 1854 贪心(分治）

Evil Straw Warts Live

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1144		Accepted: 330

Description

A palindrome is a string of symbols that is equal to itself when reversed. Given an input string, not necessarily a palindrome, compute the number of swaps necessary to transform the string into a palindrome. By swap we mean reversing the order of two adjacent symbols. For example, the string "mamad" may be transformed into the palindrome "madam" with 3 swaps: 
swap "ad" to yield "mamda" 
swap "md" to yield "madma" 
swap "ma" to yield "madam" 

Input

The first line of input gives n, the number of test cases. For each test case, one line of input follows, containing a string of up to 8000 lowercase letters.

Output

Output consists of one line per test case. This line will contain the number of swaps, or "Impossible" if it is not possible to transform the input to a palindrome. 

Sample Input

3
mamad
asflkj
aabb

Sample Output

3
Impossible
2

Source

Waterloo local 2004.09.19

题意：给你一个字符串，求最少交换相邻字符多少次能将它变为回文串。

思路：先处理两边的，然后删掉，递归操作。

代码：

 1 #include"cstdio"
 2 #include"map"
 3 #include"set"
 4 #include"cmath"
 5 #include"queue"
 6 #include"vector"
 7 #include"string"
 8 #include"cstring"
 9 #include"ctime"
10 #include"iostream"
11 #include"cstdlib"
12 #include"algorithm"
13 #define db double
14 #define ll long long
15 #define vec vector<ll>
16 #define mt  vector<vec>
17 #define ci(x) scanf("%d",&x)
18 #define cd(x) scanf("%lf",&x)
19 #define cl(x) scanf("%lld",&x)
20 #define pi(x) printf("%d\n",x)
21 #define pd(x) printf("%f\n",x)
22 #define pl(x) printf("%lld\n",x)
23 //#define rep(i, x, y) for(int i=x;i<=y;i++)
24 #define rep(i, n) for(int i=0;i<n;i++)
25 const int N   = 1e6 + 5;
26 const int mod = 1e9 + 7;
27 const int MOD = mod - 1;
28 const int inf = 0x3f3f3f3f;
29 const db  PI  = acos(-1.0);
30 const db  eps = 1e-10;
31 using namespace std;
32 int ans;
33 int c[26];
34 char s[8005];
35 void swap(char &e,char &f){
36     char ch;
37     ch=e,e=f,f=ch;
38 }
39 int dfs(int l,int r)
40 {
41     if(l>=r) return ans;
42     int lp=N,rp=-N;
43     for(int i=l;i<r;i++){//移动字符到左边与右边配对
44         if(s[i]==s[r]){
45             lp=i;
46             break;
47         }
48     }
49     for(int i=r;i>l;i--){//移动字符到右边与左边配对
50         if(s[i]==s[l]){
51             rp=i;
52             break;
53         }
54     }
55     if(lp-l<=r-rp){
56         ans+=lp-l;
57         for(int i=lp;i>l;i--) swap(s[i],s[i-1]);
58     }
59     else{
60         ans+=r-rp;
61         for(int i=rp;i<r;i++) swap(s[i],s[i+1]);
62     }
63     return dfs(l+1,r-1);
64 }
65 int main()
66 {
67     int n;
68     ci(n);
69     while(n--)
70     {
71         cin>>s;
72         memset(c,0, sizeof(c));
73         int len=strlen(s);
74         for(int i=0;s[i];i++) c[s[i]-'a']++;
75         int ok=0;
76         for(int i=0;i<26;i++) if(c[i]&1) ok++;
77         ans=0;
78         if(ok>1) puts("Impossible");
79         else pi(dfs(0,len-1));
80     }
81 }