HDU 3341 Lost's revenge （AC自动机 + DP + 变进制/hash）题解

题意：给你些分数串，给你一个主串，主串每出现一个分数串加一分，要你重新排列主串，最多几分

思路：显然这里开$40^4$去状压内存不够。但是我们自己想想会发现根本不用开那么大，因为很多状态是废状压，不是不存在的，那么可以考虑想办法简化状态。

一个是hash，直接打表所有子情况，用ha[][][][]表示出所有情况，那么直接dp[status][size]去dp。

还有一种用变进制：

假设ACGT的总数分别为num[0],num[1],num[2],num[3]

那么对于ACGT的数量分别为ABCD的状态可以记录为：

 A*(num[1]+1)*(num[2]+1)*(num[3]+1) + B*(num[2]+1)*(num[3]+1)+ C*(num[3]+1) +D

显然末尾D基数为1，次末尾C基数(num[3]+1)，那么D不管怎么变（最多变num[3]）永远影响不到C那个级数（（num[3] + 1 ）* k），那么就能区分变得是哪一位。

代码：

/*变进制*/
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 500 + 5;
const int M = 50 + 5;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 20090717;
int n, m;
int dp[15000][maxn];
int getid(char s){
    if(s == 'A') return 0;
    if(s == 'C') return 1;
    if(s == 'G') return 2;
    return 3;
}
struct Aho{
    struct state{
        int next[4];
        int fail, cnt;
    }node[maxn];
    int size;
    queue<int> q;

    void init(){
        size = 0;
        newtrie();
        while(!q.empty()) q.pop();
    }

    int newtrie(){
        memset(node[size].next, 0, sizeof(node[size].next));
        node[size].cnt = node[size].fail = 0;
        return size++;
    }

    void insert(char *s){
        int len = strlen(s);
        int now = 0;
        for(int i = 0; i < len; i++){
            int c = getid(s[i]);
            if(node[now].next[c] == 0){
                node[now].next[c] = newtrie();
            }
            now = node[now].next[c];
        }
        node[now].cnt++;
    }

    void build(){
        node[0].fail = -1;
        q.push(0);

        while(!q.empty()){
            int u = q.front();
            q.pop();
            if(node[node[u].fail].cnt && u) node[u].cnt += node[node[u].fail].cnt;
            for(int i = 0; i < 4; i++){
                if(!node[u].next[i]){
                    if(u == 0)
                        node[u].next[i] = 0;
                    else
                        node[u].next[i] = node[node[u].fail].next[i];
                }
                else{
                    if(u == 0) node[node[u].next[i]].fail = 0;
                    else{
                        int v = node[u].fail;
                        while(v != -1){
                            if(node[v].next[i]){
                                node[node[u].next[i]].fail = node[v].next[i];
                                break;
                            }
                            v = node[v].fail;
                        }
                        if(v == -1) node[node[u].next[i]].fail = 0;
                    }
                    q.push(node[u].next[i]);
                }
            }
        }
    }

    void query(char *s){
        int ans = 0;
        int len = strlen(s);
        int num[4] = {0};
        for(int i = 0; i < len; i++){
            num[getid(s[i])]++;
        }
        memset(dp, -1, sizeof(dp));
        dp[0][0] = 0;
        int fac[4];
        fac[0] = (num[1] + 1) * (num[2] + 1) * (num[3] + 1);
        fac[1] = (num[2] + 1) * (num[3] + 1);
        fac[2] = (num[3] + 1);
        fac[3] = 1;
        for(int i = 0; i <= num[0]; i++){
            for(int j = 0; j <= num[1]; j++){
                for(int k = 0; k <= num[2]; k++){
                    for(int l = 0; l <= num[3]; l++){
                        int id = i * fac[0] + j * fac[1] + k * fac[2] + l;
                        for(int g = 0; g < size; g++){
                            if(dp[id][g] == -1) continue;
                            for(int h = 0; h < 4; h++){
                                int nex;
                                if(h == 0 && i == num[0]) continue;
                                if(h == 1 && j == num[1]) continue;
                                if(h == 2 && k == num[2]) continue;
                                if(h == 3 && l == num[3]) continue;
                                nex = id + fac[h];
                                int add = node[node[g].next[h]].cnt;
                                if(dp[nex][node[g].next[h]] < dp[id][g] + add){
                                    dp[nex][node[g].next[h]] = dp[id][g] + add;
                                    if(i + j + k + l + 1 == len) ans = max(ans , dp[nex][node[g].next[h]]);
                                }
                            }
                        }
                    }
                }
            }
        }
        printf("%d
", ans);
    }

}ac;
char s[45];
int main(){
    int ca = 1;
    while(~scanf("%d", &n) && n){
        ac.init();
        for(int i = 0; i < n; i++){
            scanf("%s", s);
            ac.insert(s);
        }
        ac.build();
        scanf("%s", s);
        printf("Case %d: ", ca++);
        ac.query(s);
    }
    return 0;
}

/*Hash*/
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 500 + 5;
const int M = 50 + 5;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 20090717;
int n, m;
int dp[15000][maxn];
int ha[42][42][42][42];
int getid(char s){
    if(s == 'A') return 0;
    if(s == 'C') return 1;
    if(s == 'G') return 2;
    return 3;
}
struct Aho{
    struct state{
        int next[4];
        int fail, cnt;
    }node[maxn];
    int size;
    queue<int> q;

    void init(){
        size = 0;
        newtrie();
        while(!q.empty()) q.pop();
    }

    int newtrie(){
        memset(node[size].next, 0, sizeof(node[size].next));
        node[size].cnt = node[size].fail = 0;
        return size++;
    }

    void insert(char *s){
        int len = strlen(s);
        int now = 0;
        for(int i = 0; i < len; i++){
            int c = getid(s[i]);
            if(node[now].next[c] == 0){
                node[now].next[c] = newtrie();
            }
            now = node[now].next[c];
        }
        node[now].cnt++;
    }

    void build(){
        node[0].fail = -1;
        q.push(0);

        while(!q.empty()){
            int u = q.front();
            q.pop();
            if(node[node[u].fail].cnt && u) node[u].cnt += node[node[u].fail].cnt;    //attention
            for(int i = 0; i < 4; i++){
                if(!node[u].next[i]){
                    if(u == 0)
                        node[u].next[i] = 0;
                    else
                        node[u].next[i] = node[node[u].fail].next[i];
                }
                else{
                    if(u == 0) node[node[u].next[i]].fail = 0;
                    else{
                        int v = node[u].fail;
                        while(v != -1){
                            if(node[v].next[i]){
                                node[node[u].next[i]].fail = node[v].next[i];
                                break;
                            }
                            v = node[v].fail;
                        }
                        if(v == -1) node[node[u].next[i]].fail = 0;
                    }
                    q.push(node[u].next[i]);
                }
            }
        }
    }

    void query(char *s){
        int ans = 0;
        int len = strlen(s);
        int tot = 0;
        int num[4] = {0};
        for(int i = 0; i < len; i++){
            num[getid(s[i])]++;
        }
        for(int i = 0; i <= num[0]; i++){
            for(int j = 0; j <= num[1]; j++){
                for(int k = 0; k <= num[2]; k++){
                    for(int l = 0; l <= num[3]; l++){
                        ha[i][j][k][l] = tot;
                        for(int g = 0; g < size; g++){
                            dp[tot][g] = -1;
                        }
                        tot++;
                    }
                }
            }
        }
        for(int i = 0; i <= num[0]; i++){
            for(int j = 0; j <= num[1]; j++){
                for(int k = 0; k <= num[2]; k++){
                    for(int l = 0; l <= num[3]; l++){

                    }
                }
            }
        }
        dp[0][0] = 0;
        for(int i = 0; i <= num[0]; i++){
            for(int j = 0; j <= num[1]; j++){
                for(int k = 0; k <= num[2]; k++){
                    for(int l = 0; l <= num[3]; l++){
                        int id = ha[i][j][k][l];
                        for(int g = 0; g < size; g++){
                            if(dp[id][g] == -1) continue;
                            for(int h = 0; h < 4; h++){
                                int nex;
                                if(h == 0 && i == num[0]) continue;
                                if(h == 1 && j == num[1]) continue;
                                if(h == 2 && k == num[2]) continue;
                                if(h == 3 && l == num[3]) continue;
                                if(h == 0) nex = ha[i + 1][j][k][l];
                                if(h == 1) nex = ha[i][j + 1][k][l];
                                if(h == 2) nex = ha[i][j][k + 1][l];
                                if(h == 3) nex = ha[i][j][k][l + 1];
                                int add = node[node[g].next[h]].cnt;
                                if(dp[nex][node[g].next[h]] < dp[id][g] + add){
                                    dp[nex][node[g].next[h]] = dp[id][g] + add;
                                    if(i + j + k + l + 1 == len) ans = max(ans , dp[nex][node[g].next[h]]);
                                }
                            }
                        }
                    }
                }
            }
        }
        printf("%d
", ans);
    }

}ac;
char s[45];
int main(){
    int ca = 1;
    while(~scanf("%d", &n) && n){
        ac.init();
        for(int i = 0; i < n; i++){
            scanf("%s", s);
            ac.insert(s);
        }
        ac.build();
        scanf("%s", s);
        printf("Case %d: ", ca++);
        ac.query(s);
    }
    return 0;
}