UVA 11584 Partitioning by Palindromes

题意：把字符串划分成尽量少的回文串。

dp[i] = max{dp[j-1] + 1 | str[j....i]为回文串}。

View Code 
 1 /*
 2 Author:Zhaofa Fang
 3 Lang:C++
 4 */
 5 #include <cstdio>
 6 #include <cstdlib>
 7 #include <sstream>
 8 #include <iostream>
 9 #include <cmath>
10 #include <cstring>
11 #include <algorithm>
12 #include <string>
13 #include <utility>
14 #include <vector>
15 #include <queue>
16 #include <stack>
17 #include <map>
18 #include <set>
19 using namespace std;
20 
21 typedef long long ll;
22 #define DEBUG(x) cout<< #x << ':' << x << endl
23 #define REP(i,n) for(int i=0;i < (n);i++)
24 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)
25 #define PII pair<int,int>
26 #define PB push_back
27 #define MP make_pair
28 #define FI first
29 #define SE second
30 #define lowbit(x) (x&(-x))
31 #define INF (1<<30)
32 
33 char str[1010];
34 int dp[1010];
35 
36 bool check(char s[],int l,int r)
37 {
38     for(int i=l,j=r;i<=r;i++,j--)
39         if(s[i] != s[j])return false;
40     return true;
41 }
42 int main()
43 {
44 #ifndef ONLINE_JUDGE
45     freopen("in","r",stdin);
46 #endif
47     int T;
48     scanf("%d",&T);
49     while(T--)
50     {
51         scanf("%s",str+1);
52         int len = strlen(str+1);
53         REP(i,len+1)dp[i] = INF;
54         dp[0] = 0;
55         FOR(i,1,len)
56         {
57             FOR(j,1,i)
58             {
59                 if(check(str,j,i))
60                 {
61                     dp[i] = min(dp[i],dp[j-1] + 1);
62                 }
63             }
64         }
65         printf("%d\n",dp[len]);
66     }
67     return 0;
68 }
posted @ 2012-11-08 22:58  發_  阅读(475)  评论(0)  编辑  收藏  举报