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• # HDU 3247 Resource Archiver (AC自动机+BFS+状压DP)

题意：给定 n 个文本串，m个病毒串，文本串重叠部分可以合并，但合并后不能含有病毒串，问所有文本串合并后最短多长。

析：先把所有的文本串和病毒都插入到AC自动机上，不过标记不一样，可以给病毒标记-1，如果访问知道就知道不可行的，然后处理出两两串叠加的最小长度，这个要用bfs，在AC自动机上把这个处理出来，然后剩下的就是一个简单的DP了，dp[s][i] 表示状态为 s 时，i 串在后面，长度最短是多少。

代码如下：

```#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int maxm = 1e6 + 5;
const int mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
const int maxnode = 6e4 + 100;
const int sigma = 2;
int dist[15][15], last[15], cnt;

struct Aho{
int ch[maxnode][sigma], f[maxnode];
int val[maxnode];
int sz;

void clear(){ sz = 1;  ms(ch[0], 0); }
inline int idx(char ch){ return ch - '0'; }

void insert(char *s, int v){
int u = 0;
for(int i = 0; s[i]; ++i){
int c = idx(s[i]);
if(!ch[u][c]){
ms(ch[sz], 0);
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
if(v > 0)  val[u] |= 1<<v;
else val[u] = v;
}

void getFail(){
queue<int> q;
f[0] = 0;
for(int c = 0; c < sigma; ++c){
int u = ch[0][c];
if(u){ q.push(u);  f[u] = 0; }
}

while(!q.empty()){
int r = q.front();  q.pop();
for(int c = 0; c < sigma; ++c){
int u = ch[r][c];
if(!u){ ch[r][c] = ch[f[r]][c];  continue; }
q.push(u);
int v = f[r];
while(v && !ch[v][c])  v = f[v];
f[u] = ch[v][c];
if(val[u] > 0 && val[f[u]] > 0)  val[u] |= val[f[u]];
}
}
}

int d[maxnode];

void bfs(int s){
queue<int> q;
ms(d, INF);  d[last[s]] = 0;
q.push(last[s]);

while(!q.empty()){
int u = q.front();  q.pop();
for(int c = 0; c < sigma; ++c){
int nxt = ch[u][c];
if(d[nxt] > d[u] + 1 && val[nxt] >= 0){
d[nxt] = d[u] + 1;
q.push(nxt);
}
}
}
for(int i = 0; i < cnt; ++i)
dist[s][i] = d[last[i]];
}
};

Aho aho;

char s[500006];
int dp[2100][13];

int main(){
while(scanf("%d %d", &n, &m) == 2 && n+m){
aho.cl;
for(int i = 1; i <= n; ++i){
scanf("%s", s);
aho.insert(s, i);
}
for(int i = 0; i < m; ++i){
scanf("%s", s);
aho.insert(s, -1);
}
aho.getFail();
cnt = 1;
for(int i = 0; i < aho.sz; ++i)
if(aho.val[i] > 0)  last[cnt++] = i;
for(int i = 0; i < cnt; ++i)  aho.bfs(i);
ms(dp, INF);
dp[1][0] = 0;
int all = 1<<cnt;
FOR(i, 0, all)  for(int j = 0; j < cnt; ++j){
if(dp[i][j] == INF)  continue;
for(int k = 0; k < cnt; ++k){
if(i&1<<k)  continue;
dp[i|1<<k][k] = min(dp[i|1<<k][k], dp[i][j] + dist[j][k]);
}
}
int ans = INF;
for(int i = 0; i < cnt; ++i)
ans = min(ans, dp[all-1][i]);
printf("%d
", ans);
}
return 0;
}
```

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• 原文地址：https://www.cnblogs.com/dwtfukgv/p/7689415.html