HDU 1080 Human Gene Functions

来源：互联网发布：炫踪网络 ceo 编辑：程序博客网时间：2024/05/16 14:01

读完这个题一直以为添加空格使得两个字符串一样长。然后求最大相似度。。

做了半天没做出来。又去读了遍题。看到了 if necessary, in appropriate positions of the genes to make them equally long

然后醉了。。知道之后就比较好做了。 d【i】【j】代表第一个字符串用了 i个字符，第二个字符串用了 j个字符之后的相似度

比较简单了。因为对于某一个字符要不和空格匹配要不和字符匹配。取一个最大值就好了。

#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <cmath>#include <cstdlib>#include <string>#include <map>#include <vector>#include <set>#include <queue>#include <stack>#include <cctype>using namespace std;typedef long long LL;typedef unsigned long long ULL;#define MAXN 100+10#define INF (1<<30)#define mod 123456789int cal(char c){    if(c == 'A') return 0;    if(c == 'C') return 1;    if(c == 'G') return 2;    if(c == 'T') return 3;    if(c == ' ') return 4;}int math[MAXN][MAXN] = {{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 main (){    int t;    scanf("%d",&t);    while(t--){        int len1,len2;        char s1[MAXN],s2[MAXN];        scanf("%d%s",&len1,s1+1);        scanf("%d%s",&len2,s2+1);        int d[MAXN][MAXN] = {0};        for(int i = 0; i <= len1; i++){            for(int j = 0; j <= len2; j++){                if(i == 0 && j == 0)                    continue;                if(i == 0)                    d[i][j] = d[i][j-1] + math[4][cal(s2[j])];                else if(j == 0)                    d[i][j] = d[i-1][j] + math[cal(s1[i])][4];                else                    d[i][j] = max(d[i-1][j]+math[cal(s1[i])][4], max(d[i][j-1]+math[4][cal(s2[j])], d[i-1][j-1]+math[cal(s1[i])][cal(s2[j])]));                //printf("%d %d %d\n",i,j,d[i][j]);            }        }        printf("%d\n",d[len1][len2]);    }    return 0;}

0 0