【AC自动机】 HDOJ 3341 Lost's revenge

AC自动机+状态压缩DP。。dp过程很容易想到。。但是状态不容易压缩，一个简单的想法是开个4维数组记录所有情况，但是显然空间开不下。。。所以我们需要找一个hash函数，进行状态的压缩。。。这里用变进制来进行hash，就像秒，分钟，小时。。那样子，注意一下dp过程考虑可达不可达，还有一个数取完不可以再取这个数。。。或者变进制每个数+2，就可以不考虑这个数取完的情况。

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <climits>
#define maxn 505
#define eps 1e-6
#define mod 10007
#define INF 99999999
#define lowbit(x) ((x)&(-(x)))
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
using namespace std;

struct trie
{
int next[maxn][4];
int fail[maxn];
int end[maxn];
char s[maxn];
queue<int> q;
int hash[26];
int top, now, root;

int newnode(void)
{
end[top] = 0;
fail[top] = -1;
for(int i = 0; i < 4; i++)
next[top][i] = -1;
return top++;
}
void init(void)
{
top = 0;
root = newnode();
memset(hash, 0 ,sizeof hash);
hash['A' - 'A'] = 0;
hash['C' - 'A'] = 1;
hash['G' - 'A'] = 2;
hash['T' - 'A'] = 3;
}
void insert(void)
{
int len = strlen(s), i, k;
now = root;
for(i = 0; i < len; i++) {
k = hash[s[i] - 'A'];
if(next[now][k] == -1)
next[now][k] = newnode();
now = next[now][k];
}
end[now] += 1;
}
void build(void)
{
now = root;
for(int i = 0; i < 4; i++)
if(next[now][i] == -1)
next[now][i] = root;
else {
fail[next[now][i]] =root;
q.push(next[now][i]);
}
while(!q.empty()) {
now = q.front();
q.pop();
if(end[fail[now]]) end[now] += end[fail[now]];
for(int i = 0; i < 4; i++)
if(next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else {
fail[next[now][i]] = next[fail[now]][i];
q.push(next[now][i]);
}
}
}
}tmp;
int dp[21000][maxn];
int hash[maxn];
int h[maxn];
char s[maxn];
int n;

void init(void)
{
memset(dp, -1, sizeof dp);
memset(hash, 0, sizeof hash);
memset(h, 0, sizeof h);
}
void read(void)
{
int i;
tmp.init();
for(i = 0; i < n; i++) {
scanf("%s", tmp.s);
tmp.insert();
}
tmp.build();
scanf("%s", s);
n = strlen(s);
for(i = 0; i < n; i++)
hash[tmp.hash[s[i] - 'A']]++;
/*
h[3] = 1;
h[2] = hash[3]+1;
h[1] = (hash[3]+1)*(hash[2]+1);
h[0] = (hash[3]+1)*(hash[2]+1)*(hash[1]+1);
*/
h[0] = 1;
h[1] = hash[0]+1;
h[2] = (hash[0]+1)*(hash[1]+1);
h[3] = (hash[0]+1)*(hash[1]+1)*(hash[2]+1);
}
inline int HASH(int a, int b, int c, int d)
{
return a*h[0]+b*h[1]+c*h[2]+d*h[3];
}
void work(void)
{
int i1, i2, i3, i4, i, j, k, ans, temp;
dp[0][0] = 0;
for(i1 = 0; i1 <= hash[0]; i1++)
for(i2 = 0; i2 <= hash[1]; i2++)
for(i3 = 0; i3 <= hash[2]; i3++)
for(i4 = 0; i4 <= hash[3]; i4++){
temp = HASH(i1, i2, i3, i4);
for(j = 0; j < tmp.top; j++)
if(~dp[temp][j])
for(k = 0; k < 4; k++) {
if(k == 0 && i1 == hash[0]) continue;
if(k == 1 && i2 == hash[1]) continue;
if(k == 2 && i3 == hash[2]) continue;
if(k == 3 && i4 == hash[3]) continue;
dp[temp+h[k]][tmp.next[j][k]] = max(dp[temp+h[k]][tmp.next[j][k]], dp[temp][j] + tmp.end[tmp.next[j][k]]);
}
}
ans = 0;
temp = HASH(hash[0], hash[1], hash[2], hash[3]);
for(i = 0; i < tmp.top; i++)
ans = max(ans, dp[temp][i]);
printf("%d\n", ans);
}
int main(void)
{
int _ = 0;
while(scanf("%d", &n), n!=0) {
init();
read();
printf("Case %d: ", ++_);
work();
}
return 0;
}

变进制+2的情况

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <climits>
#define maxn 505
#define eps 1e-6
#define mod 10007
#define INF 99999999
#define lowbit(x) ((x)&(-(x)))
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
using namespace std;

struct trie
{
int next[maxn][4];
int fail[maxn];
int end[maxn];
char s[maxn];
queue<int> q;
int hash[26];
int top, now, root;

int newnode(void)
{
end[top] = 0;
fail[top] = -1;
for(int i = 0; i < 4; i++)
next[top][i] = -1;
return top++;
}
void init(void)
{
top = 0;
root = newnode();
memset(hash, 0 ,sizeof hash);
hash['A' - 'A'] = 0;
hash['C' - 'A'] = 1;
hash['G' - 'A'] = 2;
hash['T' - 'A'] = 3;
}
void insert(void)
{
int len = strlen(s), i, k;
now = root;
for(i = 0; i < len; i++) {
k = hash[s[i] - 'A'];
if(next[now][k] == -1)
next[now][k] = newnode();
now = next[now][k];
}
end[now] += 1;
}
void build(void)
{
now = root;
for(int i = 0; i < 4; i++)
if(next[now][i] == -1)
next[now][i] = root;
else {
fail[next[now][i]] =root;
q.push(next[now][i]);
}
while(!q.empty()) {
now = q.front();
q.pop();
if(end[fail[now]]) end[now] += end[fail[now]];
for(int i = 0; i < 4; i++)
if(next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else {
fail[next[now][i]] = next[fail[now]][i];
q.push(next[now][i]);
}
}
}
}tmp;
int dp[21000][maxn];
int hash[maxn];
int h[maxn];
char s[maxn];
int n;

void init(void)
{
memset(dp, -1, sizeof dp);
memset(hash, 0, sizeof hash);
memset(h, 0, sizeof h);
}
void read(void)
{
int i;
tmp.init();
for(i = 0; i < n; i++) {
scanf("%s", tmp.s);
tmp.insert();
}
tmp.build();
scanf("%s", s);
n = strlen(s);
for(i = 0; i < n; i++)
hash[tmp.hash[s[i] - 'A']]++;
/*
h[3] = 1;
h[2] = hash[3]+1;
h[1] = (hash[3]+1)*(hash[2]+1);
h[0] = (hash[3]+1)*(hash[2]+1)*(hash[1]+1);
*/
h[0] = 1;
h[1] = hash[0]+2;
h[2] = (hash[0]+2)*(hash[1]+2);
h[3] = (hash[0]+2)*(hash[1]+2)*(hash[2]+2);
}
inline int HASH(int a, int b, int c, int d)
{
return a*h[0]+b*h[1]+c*h[2]+d*h[3];
}
void work(void)
{
int i1, i2, i3, i4, i, j, k, ans, temp;
dp[0][0] = 0;
for(i1 = 0; i1 <= hash[0]; i1++)
for(i2 = 0; i2 <= hash[1]; i2++)
for(i3 = 0; i3 <= hash[2]; i3++)
for(i4 = 0; i4 <= hash[3]; i4++){
temp = HASH(i1, i2, i3, i4);
for(j = 0; j < tmp.top; j++)
if(~dp[temp][j])
for(k = 0; k < 4; k++)
dp[temp+h[k]][tmp.next[j][k]] = max(dp[temp+h[k]][tmp.next[j][k]], dp[temp][j] + tmp.end[tmp.next[j][k]]);
}
ans = 0;
temp = HASH(hash[0], hash[1], hash[2], hash[3]);
for(i = 0; i < tmp.top; i++)
ans = max(ans, dp[temp][i]);
printf("%d\n", ans);
}
int main(void)
{
int _ = 0;
while(scanf("%d", &n), n!=0) {
init();
read();
printf("Case %d: ", ++_);
work();
}
return 0;
}