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40%的数据满足:1<=n<=10
100%的数据满足:1<=n<=50
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 6 #define maxn 52 7 8 using namespace std; 9 10 char s[maxn]; 11 12 int f[maxn][maxn],inf=99999999; 13 14 int main() 15 { 16 scanf("%s",s); 17 memset(f,128/3,sizeof(f)); 18 int n=strlen(s); 19 for(int i=1;i<=n;i++) 20 f[i][i]=1; 21 for(int i=n-1;i>=1;i--) 22 for(int j=i+1;j<=n;j++) 23 if(s[i-1]==s[j-1]) 24 f[i][j]=min(f[i+1][j-1]+1,min(f[i][j-1],f[i+1][j])); 25 else 26 for(int k=i;k<j;k++) 27 f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]); 28 printf("%d",f[1][n]); 29 return 0; 30 }
【DP】BZOJ 1260: [CQOI2007]涂色paint
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原文地址:http://www.cnblogs.com/tuigou/p/4881279.html