标签：des style blog color io os ar for sp
Description
Input
Output
Sample Input
3 AC CG GT CGAT 1 AA AAA 0
Sample Output
Case 1: 3 Case 2: 2
思路：假设ACGT的个数依次是num[0],num[1],num[2],num[3],将这四个数压缩成一个数，减少内存，然后动归dp[i][status]表示到自动机上i状态此时ACGT个数的状态时的最少
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
struct Trie {
int nxt[510][4], fail[510], end[510];
int root, cnt;
int newNode() {
for (int i = 0; i < 4; i++)
nxt[cnt][i] = -1;
end[cnt++] = 0;
return cnt - 1;
}
void init() {
cnt = 0;
root = newNode();
}
int getch(char ch) {
if (ch == 'A') return 0;
else if (ch == 'C') return 1;
else if (ch == 'G') return 2;
return 3;
}
void insert(char str[]) {
int len = strlen(str);
int now = root;
for (int i = 0; i < len; i++) {
if (nxt[now][getch(str[i])] == -1)
nxt[now][getch(str[i])] = newNode();
now = nxt[now][getch(str[i])];
}
end[now]++;
}
void build() {
queue<int> q;
fail[root] = root;
for (int i = 0; i < 4; i++) {
if (nxt[root][i] == -1)
nxt[root][i] = root;
else {
fail[nxt[root][i]] = root;
q.push(nxt[root][i]);
}
}
while (!q.empty()) {
int now = q.front();
q.pop();
end[now] += end[fail[now]]; //好像每次都是这里
for (int i = 0; i < 4; i++) {
if (nxt[now][i] == -1)
nxt[now][i] = nxt[fail[now]][i];
else {
fail[nxt[now][i]] = nxt[fail[now]][i];
q.push(nxt[now][i]);
}
}
}
}
int dp[510][11*11*11*11+5];
int bit[4], num[4];
int solve(char str[]) {
int len = strlen(str);
memset(num, 0, sizeof(num));
for (int i = 0; i < len; i++)
num[getch(str[i])]++;
bit[0] = (num[1]+1) * (num[2]+1) * (num[3]+1);
bit[1] = (num[2]+1) * (num[3]+1);
bit[2] = (num[3]+1);
bit[3] = 1;
memset(dp, -1, sizeof(dp));
dp[root][0] = 0;
for (int A = 0; A <= num[0]; A++)
for (int B = 0; B <= num[1]; B++)
for (int C = 0; C <= num[2]; C++)
for (int D = 0; D <= num[3]; D++) {
int s = A * bit[0] + B * bit[1] + C * bit[2] + D * bit[3];
for (int i = 0; i < cnt; i++)
if (dp[i][s] >= 0) {
for (int k = 0; k < 4; k++) {
if (k == 0 && A == num[0]) continue;
if (k == 1 && B == num[1]) continue;
if (k == 2 && C == num[2]) continue;
if (k == 3 && D == num[3]) continue;
dp[nxt[i][k]][s+bit[k]] = max(dp[nxt[i][k]][s+bit[k]], dp[i][s]+end[nxt[i][k]]);
}
}
}
int ans = 0;
int status = num[0] * bit[0] + num[1] * bit[1] + num[2] * bit[2] + num[3] * bit[3];
for (int i = 0; i < cnt; i++)
ans = max(ans, dp[i][status]);
return ans;
}
} ac;
char str[50];
int main() {
int n, cas = 1;
while (scanf("%d", &n) != EOF && n) {
ac.init();
for (int i = 0; i < n; i++) {
scanf("%s", str);
ac.insert(str);
}
ac.build();
scanf("%s", str);
printf("Case %d: %d\n", cas++, ac.solve(str));
}
return 0;
}
HDU - 3341 Lost's revenge(AC自动机+DP)
标签：des style blog color io os ar for sp
原文地址：http://blog.csdn.net/u011345136/article/details/40185135
