poj 1080 Human Gene Functions 解题报告（附详细分析）

动态规划题，是最长子序列的变形。

最优解是字符串对应值的最大值，只需要改一下最长子序列的动态转移方程即可。

解题思路
1.取字符i-1和j-1的时候dp[i][j]=dp[i-1][j-1]+value[s1[i-1]][s2[j-1]];
2.取字符i-1不取j-1的时候dp[i][j]=dp[i-1][j]+value[s1[i-1]]['-'];
3.取字符j-1不取i-1的时候dp[i][j]=dp[i][j-1]+value['-'][s2[j-1]];

#include <iostream>using namespace std;const int MAX = 200;int max(int, int, int);int i, j;int dp[MAX][MAX];char s1[MAX], s2[MAX];int value[MAX][MAX];int main(){value['A']['A']=value['C']['C']=value['G']['G']=value['T']['T']=5;    value['A']['C']=value['C']['A']=value['A']['T']=value['T']['A']=-1;    value['-']['T']=value['T']['-']=-1;    value['A']['G']=value['G']['A']=value['C']['T']=value['T']['C']=-2;    value['G']['T']=value['T']['G']=value['G']['-']=value['-']['G']=-2;    value['A']['-']=value['-']['A']=value['C']['G']=value['G']['C']=-3;    value['C']['-']=value['-']['C']=-4;//录入题目的表格int t, m, n;scanf("%d", &t);while (t--){scanf("%d%s", &n, s1+1);scanf("%d%s", &m, s2+1);memset(dp, 0, sizeof(dp));dp[0][0] = 0;for(i = 1; i <= n; i++)dp[i][0] = dp[i-1][0] + value[s1[i]]['-'];//数组边界的初始化for(j = 1; j <= m; j++)dp[0][j] = dp[0][j-1] + value['-'][s2[j]];//当i=0时，表示第j个字母和‘-’匹配             for(i = 1; i <= n; i++)for(j = 1; j <= m; j++){dp[i][j] = max( dp[i-1][j-1] + value[s1[i]][s2[j]],dp[i][j-1] + value['-'][s2[j]],//表示s1[i]与s2[j-1]匹配，于是s2[j]就只能与‘-’匹配dp[i-1][j] + value[s1[i]]['-'] );//表示 s1[i-1] 与 s2[j] 匹配 于是 s1[i] 就只能和 ‘-’匹配}   printf("%d\n", dp[n][m]);}return 0;}int max(int a, int b, int c){int m;if(a > b)m = a;elsem = b;if(m < c)m = c;return m;}