hdu 1080 Human Gene Functions 简单dp

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传送门　http://acm.hdu.edu.cn/showproblem.php?pid=1080

题意:
- 给出两串 dna 序列, 每个碱基之间匹配得分不同要求匹配出最大得分

分析:
- dp[i][j] 代表匹配到第一串第 i 个和第二串第 j 个时的最大得分
- dp[i][j] = max(dp[i-1][j-1] + match(i, j), dp[i][j-1] - match(i, j-1) + match(0, j-1) + match(i, j), dp[i-1][j] - match(i-1, j) + match(i-1, 0) + match(i, j));

#include <stdio.h>#include <iostream>#include <cstring>#include <string>using namespace std;const int MAX = 111, INF = 1 << 16 - 1;const int mch[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 dp[MAX][MAX], mp[2][MAX], T, n[2];char c;int max(int a, int b) {return a > b ? a : b;}int judge(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;    return -1;}void getin() {    for (int i = 0; i < 2; ++i) {        scanf ("%d", n+i);        c = 0;        while (true) {            c = getchar();            if (c == 'A' || c == 'T' || c == 'G' || c == 'C')                 break;        }        mp[i][1] = judge(c);        for (int j = 2; j <= n[i]; ++j) {            c = getchar();            mp[i][j] = judge(c);        }    }    for (int i = 0; i < MAX; ++i)        for (int j = 0; j < MAX; ++j)            dp[i][j] = -INF;    dp[0][0] = 0;}void DP() {    for (int i = 0; i <= n[0]; ++i) {        for (int j = 0; j <= n[1]; ++j) {            if (i && j)                 dp[i][j] = max(dp[i-1][j-1] + mch[mp[0][i]][mp[1][j]], dp[i][j]);            if (i)                 dp[i][j] = max(dp[i-1][j] - mch[mp[0][i-1]][mp[1][j]] +                                 mch[mp[0][i-1]][4] + mch[mp[0][i]][mp[1][j]], dp[i][j]);            if (j)                 dp[i][j] = max(dp[i][j-1] - mch[mp[0][i]][mp[1][j-1]] +                                 mch[4][mp[1][j-1]] + mch[mp[0][i]][mp[1][j]], dp[i][j]);        }    }    printf("%d\n", dp[n[0]][n[1]]);}int main() {    freopen("in.t", "r", stdin);    scanf("%d", &T);    while (T--) {        getin();        DP();    }    return 0;}

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2017-10-10

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