poj 1080 dp

dp题目最重要的是找到递推关系和表示关系， 找到这两点基本就ok了， 

如本题 dp[i][j]表示的含义第一个字符串的i个和s2的第j个匹配，它由三种情况得到，

重点是理解：dp[i][j]为取s1第i个字符，s2第j个字符时的最大分值

则决定dp为最优的情况有三种（score[][]为s1[i]和s2[j]两符号的分数）：

1、  s1取第i个字母，s2取“ - ”：由dp[i-1][j]+score[ s1[i-1] ]['-']; 推出开始不太明白， dp[i][j] = dp[i-][j] + score[s1[i-1]]['-'] 用s1[i-1]是由于

数组从0开始， 第i个字符就是字符串的s1[i-1]个，理解这个就好了

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int score[5][5] = {{0, -3, -4, -2, -1},
             {-3, 5, -1, -2, -1},
             {-4, -1, 5, -3, -2},
             {-2, -2, -3, 5, -2},
             {-1, -1, -2, -2, 5}
            };
int fen[120][120];
int s1[120], s2[120];
int n;

int exchange(char ch) {

    if(ch == 'A')
        return 1;
    else if(ch == 'C')
        return 2;
    else if(ch == 'G')
        return 3;
    else
        return 4;
}

void init() {

    memset(s1, 0, sizeof(s1));
    memset(s2, 0, sizeof(s2));
    memset(fen, 0, sizeof(fen));
}

int main()
{
    int m1, m2;
    char ch;
    int a, b, c;

    scanf("%d", &n);

    for(int i = 0; i < n; i++) {

        init();
        scanf("%d", &m1);
         getchar();
        for(int j = 0; j < m1; j++) {

            scanf("%c", &ch);
            s1[j] = exchange(ch);
            fen[s1[j]][0] = score[0][s2[j]];

        }

        scanf("%d", &m2);
        getchar();
        for(int j = 0; j < m2; j++){

            scanf("%c", &ch);
            s2[j] = exchange(ch);

        }
 //第一种写法；
        for(int i = 1; i <= m1; i++)
            fen[i][0] = fen[i-1][0] + score[0][s1[i-1]];
        for(int i = 1; i <= m2; i++)
            fen[0][i] = fen[0][i-1] + score[0][s2[i-1]];

            for(int i = 1; i <= m1; i++)
            for(int j = 1; j <= m2; j++) {


            a = fen[i-1][j-1] + score[s1[i-1]][s2[j-1]];
            b = fen[i-1][j] + score[0][s1[i-1]];
            c = fen[i][j-1] + score[0][s2[j-1]];
           fen[i][j] = max(a, max(b, c));

            }

 //我的写法
   /* for(int i = 0; i <= m1; i++)

        for(int j = 0; j <= m2; j++) {

        if(i > 0 && j > 0) {

             a = fen[i-1][j-1] + score[s1[i-1]][s2[j-1]];
            b = fen[i-1][j] + score[0][s1[i-1]];
            c = fen[i][j-1] + score[0][s2[j-1]];
           fen[i][j] = max(a, max(b, c));

        }else if(i > 0){

            fen[i][j] = fen[i-1][j] + score[0][s1[i-1]];

        }else if(j > 0) {

            fen[i][j] = fen[i][j-1] + score[0][s2[j-1]];
        }
    }*/

     printf("%d\n", fen[m1][m2]);
    }

    return 0;
}

2、  s1取“ - ”，s2取第j个字母：dp[i][j-1]+score['-'][ s2[j-1] ];

3、  s1取第i个字母，s2取第j个字母：dp[i-1][j-1]+score[ s1[i-1] ][ s2[j-1] ];

 即dp[i][j]=max( dp[i-1][j]+score[ s1[i-1] ]['-'],

dp[i][j-1]+score['-'][ s2[j-1] ],

dp[i-1][j-1]+score[ s1[i-1] ][ s2[j-1] ] );