HDU 4511 (AC自动机 DP)

题目链接:点击这里

题意:给定n个点的坐标, 每次只能从一个点走到编号比他大的点. 求一条最短的路径, 并且这条路径上不能有给出的子路径.

把这些子路径插进AC自动机里, 然后用dp[i][j]表示在图上的i, 字典树上的j的最短路径, 转移方程就是dp[i][j]=min{dp[p[k]+dis(p,i)∣∣next[k][i]=j}

#include <bits/stdc++.h>
using namespace std;
#define maxn 505
#define INF 1e20

int n, m, k;
int buf[maxn];
double p[maxn][2];

double dis (int i, int j) {
double x = p[i][0]-p[j][0], y = p[i][1]-p[j][1];
return sqrt (x*x + y*y);
}

struct trie {
int next[maxn][55], fail[maxn], end[maxn];
int root, cnt;
int new_node () {
memset (next[cnt], -1, sizeof next[cnt]);
end[cnt++] = 0;
return cnt-1;
}
void init () {
cnt = 0;
root = new_node ();
}
void insert (int *buf, int len) {//字典树插入一条路径
int now = root;
for (int i = 0; i < len; i++) {
int id = buf[i];
if (next[now][id] == -1) {
next[now][id] = new_node ();
}
now = next[now][id];
}
end[now]++;
}
void build () {//构建fail指针
queue <int> q;
fail[root] = root;
for (int i = 1; i <= n; i++) {
if (next[root][i] == -1) {
next[root][i] = root;
}
else {
fail[next[root][i]] = root;
q.push (next[root][i]);
}
}
while (!q.empty ()) {
int now = q.front (); q.pop ();
end[now] += end[fail[now]];
for (int i = 1; i <= n; 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]);
}
}
}
}
double dp[55][maxn];//在字典树上i 图上的j
void query () {
for (int i = 1; i <= n; i++) {
for (int j = 0; j < cnt; j++) {
dp[i][j] = INF;
}
}
dp[1][next[root][1]] = 0;
for (int i = 1; i < n; i++) {
for (int j = 0; j < cnt; j++) {
if (end[j] || dp[i][j] >= INF) continue;
for (int x = i+1; x <= n; x++) {
int nexti = next[j][x];
if (end[nexti]) continue;
dp[x][nexti] = min (dp[x][nexti], dp[i][j] + dis (i, x));
}
}
}

double ans = INF;
for (int i = 0; i < cnt; i++)
ans = min (ans, dp
[i]);
if (ans == INF) {
printf ("Can not be reached!\n");
}
else
printf ("%.2f\n", ans);
}
}ac;

int main () {
//freopen ("in.txt", "r", stdin);
while (cin >> n >> m && n+m) {
for (int i = 1; i <= n; i++) {
cin >> p[i][0] >> p[i][1];
}
ac.init ();
for (int i = 0; i < m; i++) {
cin >> k;
for (int j = 0; j < k; j++) {
cin >> buf[j];
}
ac.insert (buf, k);
}
ac.build ();
ac.query ();
}
return 0;
}