【UVa】Partitioning by Palindromes（dp）

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=27&page=show_problem&problem=2549

设w[i,j]为i-j能分割成的最少回文串

f[i]为前i个字符能够分成的最少回文串

w[i,j]=1 当w[i+1,j-1]==1 && s[i]==s[j] 或 i==j-1 && s[i]==s[j]

w[i,j]=w[i+1,j-1]+2 当s[i]!=s[j]

然后

f[i]=min{f[j]+w[j+1,i], 0<=j<i}

f[0]=0

题目白书有

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

using namespace std;

typedef long long ll;

#define pii pair<int, int>

#define mkpii make_pair<int, int>

#define pdi pair<double, int>

#define mkpdi make_pair<double, int>

#define pli pair<ll, int>

#define mkpli make_pair<ll, int>

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << (#x) << " = " << (x) << endl

#define error(x) (!(x)?puts("error"):0)

#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }

#define printarr1(a, b) for1(_, 1, b) cout << a[_] << '\t'; cout << endl

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }



const int N=1005;

int n, f[N], w[N][N];

char s[N];

int main() {

	int cs=getint();

	while(cs--) {

		scanf("%s", s+1);

		n=strlen(s+1);

		CC(f, 0x3f); CC(w, 0); f[0]=0;

		for1(i, 1, n) w[i][i]=1;

		for1(k, 1, n-1)

			for1(i, 1, n-k) {

				int j=i+k;

				if(k==1 && s[i]==s[j]) w[i][j]=1;

				else if(k>1 && w[i+1][j-1]==1 && s[i]==s[j]) w[i][j]=1;

				else w[i][j]=w[i+1][j-1]+2;

			}

		for1(i, 1, n) rep(j, i) f[i]=min(f[i], f[j]+w[j+1][i]);

		printf("%d\n", f[n]);

	}

	return 0;

}