BZOJ 1260&UVa 4394 区间DP

题意:

　　给一段字符串成段染色,问染成目标串最少次数.

SOL:

　　区间DP...

　　DP[i][j]表示从i染到j最小代价

　　转移:dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);

CODE:

　　BZ:

/*=================================================================
# Created time: 2016-03-28 21:10
# Filename: uva4394.cpp
# Description:
=================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <cmath>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm> 

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector> 

#define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1] 

#define maxn 500
#define maxm 100000
#define pi 3.1415926535898
#define _e 2.718281828459
#define INF 1070000000
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 

template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
/*==================split line==================*/
char A[maxn],C[maxn];
int dp[maxn][maxn],ans[maxn];
int main(){
	freopen("a.in","r",stdin);
	//while (scanf("%s",C)!=EOF){
		 scanf("%s",A);
		//FORP(i,0,maxn) B[i]='';
		int n=strlen(A);
		memset(dp,0,sizeof(dp));
		FORP(i,0,n-1)
			FORP(j,i,n-1) dp[i][j]=INF;
		FORP(i,0,n-1) dp[i][i]=1;
		 FORP(l,1,n-1)
		 	FORP(i,0,n-1-l)
			{
				int j=i+l;
				dp[i][j]=dp[i+1][j]+1;
				FORP(k,i+1,j)
				if (A[k]==A[i]) dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
			}
	printf("%d",dp[0][n-1]);
	//FORP(i,0,n-1)
		//	FORP(j,i,n-1) printf("%d%c",dp[i][j],j==n-1?'\n':' ');
		/*memset(ans,0,sizeof(ans));
		ans[0]=(A[0]!=C[0]);
		FORP(i,1,n-1)
			if (A[i]==C[i])	ans[i]=ans[i-1];
			else {
				ans[i]=dp[0][i];
				FORP(j,0,i-1) ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
			}
		printf("%d\n",ans[n-1]);
	}*/
}

UVa:

　　

/*=================================================================
# Created time: 2016-03-28 21:10
# Filename: uva4394.cpp
# Description:
=================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <cmath>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm> 

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector> 

#define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1] 

#define maxn 500
#define maxm 100000
#define pi 3.1415926535898
#define _e 2.718281828459
#define INF 1070000000
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 

template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
/*==================split line==================*/
char A[maxn],C[maxn];
int dp[maxn][maxn],ans[maxn];
int main(){
	freopen("a.in","r",stdin);
	while (scanf("%s",C)!=EOF){
		 scanf("%s",A);
		//FORP(i,0,maxn) B[i]='';
		int n=strlen(A);
		memset(dp,0,sizeof(dp));
		FORP(i,0,n-1)
			FORP(j,i,n-1) dp[i][j]=INF;
		FORP(i,0,n-1) dp[i][i]=1;
		 FORP(l,1,n-1)
		 	FORP(i,0,n-1-l)
			{
				int j=i+l;
				dp[i][j]=dp[i+1][j]+1;
				FORP(k,i+1,j)
				if (A[k]==A[i]) dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
			}
		//FORP(i,0,n-1)
		//	FORP(j,i,n-1) printf("%d%c",dp[i][j],j==n-1?'\n':' ');
		memset(ans,0,sizeof(ans));
		ans[0]=(A[0]!=C[0]);
		FORP(i,1,n-1)
			if (A[i]==C[i])	ans[i]=ans[i-1];
			else {
				ans[i]=dp[0][i];
				FORP(j,0,i-1) ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
			}
		printf("%d\n",ans[n-1]);
	}
}