POJ 3280 简单区间 DP

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就是想的时候不要想漏了转移方式。

对于一个回文串去掉最左和最右两个后依然是回文串，好了。。。dp

转移有三种，一种是从 dp[i+1][j] 转移过来，一种是从 dp[i][j-1] 转移过来，一种是从 dp[i-1][j-1] 转移过来

#include <stdio.h>#include <iostream>#include <queue>#include <algorithm>#include <map>#include <vector>#include <cmath>#include <string.h>#include <stdlib.h>#include <time.h>#include <fstream>#include <set>#include <stack>using namespace std;#define READ freopen("acm.in","r",stdin)#define WRITE freopen("acm.out","w",stdout)#define ll long long#define ull unsigned long long #define PII pair<int,int>#define PDI pair<double,int>#define PDD pair<double,double>#define MII map<int,int>::iterator #define fst first#define sec second#define MS(x,d) memset(x,d,sizeof(x))#define INF 0x3f3f3f3f#define ALL(x) x.begin(),x.end()#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ROOT 0,n-1,1#define PB push_back#define FOR(a,b,c) for(int a=b;a<c;a++)#define MOD 1000000007#define keyTree (ch[ ch[root][1] ][0])#define MAX 2222int dp[MAX][MAX];int add[100];int del[100];char s[MAX];int main(){       int n,len;    READ;    while(scanf("%d%d",&n,&len)!=EOF)    {        scanf("%s",s+1);        for(int i=0;i<n;i++)        {            char op[3];            int a,d;            scanf("%s%d%d",op,&a,&d);            add[op[0]-'a']=a;            del[op[0]-'a']=d;        }        for(int i=0;i<MAX;i++)            for(int j=i+1;j<MAX;j++)                dp[i][j]=INF;        for(int l=2;l<=len;l++)        {            for(int i=1;i+l-1<=len;i++)            {                int j=i+l-1;                if(s[i]==s[j])                    dp[i][j]=min(dp[i][j],dp[i+1][j-1]);                int idl=s[i]-'a',idr=s[j]-'a';                int cr=min(dp[i][j-1]+add[idr],dp[i][j-1]+del[idr]);                int cl=min(dp[i+1][j]+add[idl],dp[i+1][j]+del[idl]);                int cm=min(dp[i+1][j-1]+add[idl]+add[idr],dp[i+1][j-1]+del[idl]+del[idr]);                int t1=min(cr,cl);                int t2=min(dp[i][j],cm);                dp[i][j]=min(dp[i][j],t1);            }        }        cout<<dp[1][len]<<endl;    }    return 0;}

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