Resource Archiver HDU - 3247 AC自动机+BFS+状压
2024-08-28 07:48:49
题意:
给出n个资源串,m个病毒串,现在要如何连接资源串使得不含病毒串(可以重叠,样例就是重叠的)。
题解:
这题的套路和之前的很不同了,之前的AC自动机+DP的题目一般都是通过teir图去转移,
这题有点小不同,认真分析这一题,我们可以知道n个资源串,m个病毒串在AC自动机的上面是哪一个节点
所以可以根据teir图去处理出每一个资源串的节点不经过病毒串节点的最短距离,这个可以通过BFS实现。
这个一开始要处理AC自动机上的root节点,毕竟是从root节点开始走的。
处理出了每个点到其他点的最短距离之后,就变成了一个TSP问题,必须经过几个点的最短距离,状压写即可。
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map> #define pi acos(-1.0)
#define eps 1e-9
#define fi first
#define se second
#define rtl rt<<1
#define rtr rt<<1|1
#define bug printf("******\n")
#define mem(a, b) memset(a,b,sizeof(a))
#define name2str(x) #x
#define fuck(x) cout<<#x" = "<<x<<endl
#define sfi(a) scanf("%d", &a)
#define sffi(a, b) scanf("%d %d", &a, &b)
#define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c)
#define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define sfL(a) scanf("%lld", &a)
#define sffL(a, b) scanf("%lld %lld", &a, &b)
#define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c)
#define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
#define sfs(a) scanf("%s", a)
#define sffs(a, b) scanf("%s %s", a, b)
#define sfffs(a, b, c) scanf("%s %s %s", a, b, c)
#define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
#define FIN freopen("../in.txt","r",stdin)
#define gcd(a, b) __gcd(a,b)
#define lowbit(x) x&-x
#define IO iOS::sync_with_stdio(false) using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const ULL seed = ;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + ;
const int maxm = 8e6 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int n, m;
char buf[maxn]; struct Aho_Corasick {
int next[][], fail[], End[];
int root, cnt; int newnode() {
for (int i = ; i < ; i++) next[cnt][i] = -;
End[cnt++] = ;
return cnt - ;
} void init() {
cnt = ;
root = newnode();
} void insert(char buf[], int id) {
int len = strlen(buf);
int now = root;
for (int i = ; i < len; i++) {
if (next[now][buf[i] - ''] == -) next[now][buf[i] - ''] = newnode();
now = next[now][buf[i] - ''];
}
End[now] = id;
} void build() {
queue<int> Q;
fail[root] = root;
for (int i = ; i < ; i++)
if (next[root][i] == -) next[root][i] = root;
else {
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while (!Q.empty()) {
int now = Q.front();
Q.pop();
if (End[fail[now]] == -) End[now] = -;
else End[now] |= End[fail[now]];
for (int i = ; i < ; i++)
if (next[now][i] == -) next[now][i] = next[fail[now]][i];
else {
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
} int mp[][], dis[], pos[], num, dp[][]; void bfs(int x) {
mem(dis, -);
dis[pos[x]] = ;
queue<int> q;
q.push(pos[x]);
while (!q.empty()) {
int now = q.front();
q.pop();
for (int i = ; i < ; ++i) {
int idx = next[now][i];
if (End[idx] == - || dis[idx] >= ) continue;
dis[idx] = dis[now] + , q.push(idx);
}
}
for (int i = ; i < num; ++i) mp[x][i] = dis[pos[i]];
} int solve() {
pos[] = , num = ;
for (int i = ; i < cnt; ++i) if (End[i] > ) pos[num++] = i;
for (int i = ; i < num; ++i) bfs(i); // for (int i = 0; i < num; ++i) {
// for (int j = 0; j < num; ++j) {
// printf("%d ", mp[i][j]);
// }
// printf("\n");
// } for (int i = ; i < num; ++i)
for (int j = ; j < ( << n); ++j)
dp[i][j] = INF;
dp[][] = ;
for (int status = ; status < ( << n); ++status) {
for (int i = ; i < num; ++i) {
if (dp[i][status] == INF) continue;
for (int j = ; j < num; ++j) {
if (mp[i][j] == - || (status | End[pos[j]]) == status) continue;
dp[j][status | End[pos[j]]] = min(dp[j][status | End[pos[j]]], dp[i][status] + mp[i][j]);
}
}
}
int ans = INF;
for (int i = ; i < num; ++i) ans = min(ans, dp[i][( << n) - ]);
return ans;
} void debug() {
for (int i = ; i < cnt; i++) {
printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
for (int j = ; j < ; j++) printf("%2d", next[i][j]);
printf("]\n");
}
}
} ac; int main() {
// FIN;
while (~sffi(n, m)) {
if (n == && m == ) break;
ac.init();
for (int i = ; i < n; ++i) {
sfs(buf);
ac.insert(buf, ( << i));
}
for (int i = ; i < m; ++i) {
sfs(buf);
ac.insert(buf, -);
}
ac.build();
printf("%d\n", ac.solve());
}
return ;
}
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