HDU 4511 小明系列故事——女友的考验 ( Trie图 && DP )

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<math.h>
using namespace std;
const int Max_Tot = 555;
const int Letter  = 55;
const double INF = 1e20;///不能使用 0x3f3f3f3f
int n, m;
pair<int, int> Point[55];
struct Aho{
    struct StateTable{
        int Next[Letter];
        int fail, flag;
    }Node[Max_Tot];
    int Size;
    queue<int> que;

    inline void init(){
        while(!que.empty()) que.pop();
        memset(Node[0].Next, 0, sizeof(Node[0].Next));
        Node[0].fail = Node[0].flag = 0;
        Size = 1;
    }

    inline void insert(int *s, int len){
        int now = 0;
        for(int i=0; i<len; i++){
            int idx = s[i];
            if(!Node[now].Next[idx]){
                memset(Node[Size].Next, 0, sizeof(Node[Size].Next));
                Node[Size].fail = Node[Size].flag = 0;
                Node[now].Next[idx] = Size++;
            }
            now = Node[now].Next[idx];
        }
        Node[now].flag = 1;
    }

    inline void BuildFail(){
        Node[0].fail = 0;
        for(int i=0; i<n; i++){
            if(Node[0].Next[i]){
                Node[Node[0].Next[i]].fail = 0;
                que.push(Node[0].Next[i]);
            }else Node[0].Next[i] = 0;///必定指向根节点
        }
        while(!que.empty()){
            int top = que.front(); que.pop();
            Node[top].flag |= Node[Node[top].fail].flag;///注意了!!
            for(int i=0; i<n; i++){
                int &v = Node[top].Next[i];
                if(v){
                    que.push(v);
                    Node[v].fail = Node[Node[top].fail].Next[i];
                }else v = Node[Node[top].fail].Next[i];
            }
        }
    }
}ac;


double GetDis(pair<int,int> &st, pair<int,int> &en)
{
    double x1 =  (double)st.first;
    double x2 =  (double)en.first;
    double y1 =  (double)st.second;
    double y2 =  (double)en.second;
    return sqrt( (double)(1.0*x1-x2)*(double)(1.0*x1-x2) +
                 (double)(1.0*y1-y2)*(double)(1.0*y1-y2));
}

double dp[55][555];
inline void Solve()
{

    for(int i=0; i<n; i++)
        for(int j=0; j<ac.Size; j++)
            dp[i][j] = INF;

    dp[0][ac.Node[0].Next[0]] = 0;
    for(int i=0; i<n-1; i++){
        for(int j=0; j<ac.Size; j++){
            if(dp[i][j] < INF){
                for(int k=i+1; k<n; k++){
                    int newi = k;
                    int newj = ac.Node[j].Next[k];
                    if(!ac.Node[ newj ].flag){
                        dp[newi][newj] = min(dp[newi][newj],
                                             dp[i][j]+GetDis(Point[i], Point[k]));
                    }
                }
            }
        }
    }

    double ans = INF;
    for(int i=0; i<ac.Size; i++)
        if(dp[n-1][i] < INF)
            ans = min(ans, dp[n-1][i]);

    if(ans == INF) puts("Can not be reached!");
    else printf("%.2lf\n", ans);
}

int main(void)
{
    while(~scanf("%d %d", &n, &m)){
        if(n==0 && m==0) break;
        for(int i=0; i<n; i++)///这里我将编号 -1 了，也就是编号从 0 ~ n-1，所以下面操作都是按个编号来
            scanf("%d %d", &Point[i].first, &Point[i].second);
        int tmp[10];
        ac.init();
        while(m--){
            int k;
            scanf("%d", &k);
            for(int i=0; i<k; i++){
                scanf("%d", &tmp[i]);
                tmp[i]--;///因为编号是 0 ~ n-1
            }
            ac.insert(tmp, k);
        }ac.BuildFail();
        Solve();
    }
    return 0;
}