poj 1080 dp(sequence alignment)

题意：经典sequence alignment。给定两个ACGT串，求其最优距离。其中字母配对值给定。

思路：dp。

#include <stdio.h>
#include <string.h>
#define N 105
char s[N],t[N];
int dp[N][N],p[256][256],T;
int max(int a,int b,int c){
	if(a>b)
		return (a>c?a:c);
	return (b>c?b:c);
}
int main(){
	freopen("a.txt","r",stdin);
	scanf("%d",&T);
	p['A']['A'] = 5;p['A']['C'] = -1;p['A']['G'] = -2;p['A']['T'] = -1;p['A']['-'] = -3;
	p['C']['A'] = -1;p['C']['C'] = 5;p['C']['G'] = -3;p['C']['T'] = -2;p['C']['-'] = -4;
	p['G']['A'] = -2;p['G']['C'] = -3;p['G']['G'] = 5;p['G']['T'] = -2;p['G']['-'] = -2;
	p['T']['A'] = -1;p['T']['C'] = -2;p['T']['G'] = -2;p['T']['T'] = 5;p['T']['-'] = -1;
	p['-']['A'] = -3;p['-']['C'] = -4;p['-']['G'] = -2;p['-']['T'] = -1;
	while(T--){
		int i,j,a,b;
		memset(dp,0,sizeof(dp));
		scanf("%d %s",&a,s+1);
		scanf("%d %s",&b,t+1);
		for(i = 1;i<=a;i++)
			dp[i][0] = dp[i-1][0]+p[s[i]]['-'];
		for(j = 1;j<=b;j++)
			dp[0][j] = dp[0][j-1]+p[t[j]]['-'];
		for(i = 1;i<=a;i++)
			for(j = 1;j<=b;j++)
				dp[i][j]=max(dp[i-1][j-1]+p[s[i]][t[j]],dp[i-1][j]+p[s[i]]['-'],dp[i][j-1]+p['-'][t[j]]);
		printf("%d\n",dp[a][b]);
	}
	return 0;
}