zoj 1027 DP Human Gene Functions

此题满足动态规划求解条件：1. 最优子结构性质 2. 重叠子问题性质

每个位置可能有三种情况：1. 第一个序列被插入‘_' 2. 第二个序列被插入'_' 3. 两个都不插入，直接匹配

最优值的递归定义为c[i][j] = max(c[i-1][j-1] + arr[sx[i]][sy[j]], c[i][j-1] + arr[4][sy[j]], c[i-1][j] + arr[sx[i]][4])

#include <iostream>

using namespace std;

int max(int a, int b, int c) {
	int tmp = a > b? a : b;
	return tmp > c ? tmp : c;
}

int main() {
	int arr[5][5] = {{5,-1,-2,-1,-3}, {-1,5,-3,-2,-4}, {-2,-3,5,-2,-2}, {-1,-2,-2,5,-1}, {-3,-4,-2,-1,0}};
	int num;
	cin >> num;
	while(num--) {
		// sx,sy分别存储两个序列字符对应的下标，c[i][j]存储最优值 
		int sx[101], sy[101], c[101][101] = {0};
		char ch;
		int len1, len2;
		cin >> len1;
		for(int i = 1; i <= len1; i++) {
			cin >> ch;
			switch(ch) {						//将第i个字符转换成数组下标 
				case 'A' : sx[i] = 0; break;
				case 'C' : sx[i] = 1; break;
				case 'G' : sx[i] = 2; break;
				case 'T' : sx[i] = 3; break;
			}
		}
		cin >> len2;
		for(int i = 1; i <= len2; i++) {
			cin >> ch;
			switch(ch) {
				case 'A' : sy[i] = 0; break;
				case 'C' : sy[i] = 1; break;
				case 'G' : sy[i] = 2; break;
				case 'T' : sy[i] = 3; break;
			}
		}
		for(int i = 1; i <= len1; i++)
			c[i][0] = c[i-1][0] + arr[sx[i]][4];
		for(int j = 1; j <= len2; j++)
			c[0][j] = c[0][j-1] + arr[4][sy[j]];
		for(int i = 1; i <= len1; i++)  {
			for(int j = 1; j<= len2; j++)
				c[i][j] = max(c[i-1][j-1] + arr[sx[i]][sy[j]], c[i][j-1] + arr[4][sy[j]], c[i-1][j] + arr[sx[i]][4]);
		}
		cout << c[len1][len2] << endl;	
	}
	return 0;	
}