预处理出来任意两点的距离,然后可以顺着trie树中的节点走,不能走到不合法的地方,另开一维表示走到了哪里,依次来更新。
注意判断一下起点是不是合法。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 510 12 #define LL long long 13 const double eps = 1e-8; 14 const double pi = acos(-1.0); 15 const double INF = 1e21; 16 const int child_num = 51; 17 double w[55][55]; 18 class AC 19 { 20 private: 21 int ch[N][child_num]; 22 int fail[N]; 23 int val[N]; 24 int id[128]; 25 int Q[N]; 26 double dis[55][N]; 27 int sz; 28 public: 29 void init() 30 { 31 fail[0] = 0; 32 for(int i = 0 ; i< child_num ; i++) 33 id[i] = i; 34 } 35 void reset() 36 { 37 memset(ch[0],0,sizeof(ch[0])); 38 memset(val,0,sizeof(val)); 39 sz = 1; 40 } 41 void insert(int *a,int k,int key) 42 { 43 int i,p=0; 44 for(i = 1;i <= k; i++) 45 { 46 int d = a[i]; 47 if(ch[p][d]==0) 48 { 49 memset(ch[sz],0,sizeof(ch[sz])); 50 ch[p][d]= sz++; 51 } 52 p = ch[p][d]; 53 } 54 val[p] = key; 55 } 56 void construct() 57 { 58 int i,head= 0 ,tail = 0; 59 for(i = 1 ;i < child_num ; i++) 60 { 61 if(ch[0][i]) 62 { 63 fail[ch[0][i]] = 0; 64 Q[tail++] = ch[0][i]; 65 } 66 } 67 while(head!=tail) 68 { 69 int u = Q[head++]; 70 val[u]|=val[fail[u]]; 71 for(i = 1; i < child_num ; i++) 72 { 73 if(ch[u][i]) 74 { 75 fail[ch[u][i]] = ch[fail[u]][i]; 76 Q[tail++] = ch[u][i]; 77 } 78 else ch[u][i] = ch[fail[u]][i]; 79 } 80 } 81 } 82 void work(int n) 83 { 84 int i,j,g; 85 for(i = 1; i <= n ;i++) 86 for(j = 0 ; j <= sz ;j++) 87 dis[i][j] = INF; 88 if(val[1]) 89 { 90 puts("Can not be reached!"); 91 return ; 92 } 93 dis[1][ch[0][1]] = 0; 94 for(i = 1 ; i <= n ;i++) 95 { 96 for(j = 0; j < sz ;j++) 97 { 98 for(g = 1; g <= n ;g++) 99 { 100 int np = ch[j][g]; 101 if(val[np]) continue; 102 dis[g][np] = min(dis[g][np],dis[i][j]+w[i][g]); 103 } 104 } 105 } 106 double ans = INF; 107 for(i = 0 ; i < sz ; i++) 108 { 109 ans = min(ans,dis[n][i]); 110 } 111 if(fabs(ans-INF)<eps) puts("Can not be reached!"); 112 else 113 printf("%.2f\n",ans); 114 } 115 }ac; 116 struct node 117 { 118 double x,y; 119 }pp[55]; 120 double cdis(node a,node b) 121 { 122 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 123 } 124 int a[10]; 125 int main() 126 { 127 int n,m,k,i,j; 128 ac.init(); 129 while(scanf("%d%d",&n,&m)&&n&&m) 130 { 131 ac.reset(); 132 for(i = 1 ;i <= n; i++) 133 { 134 scanf("%lf%lf",&pp[i].x,&pp[i].y); 135 } 136 for(i = 1; i <= n ;i++) 137 for(j = 1; j <= n ;j++) 138 w[i][j] = cdis(pp[i],pp[j]); 139 for(i = 1; i <= m ; i++) 140 { 141 scanf("%d",&k); 142 for(j= 1 ;j <= k; j++) 143 scanf("%d",&a[j]); 144 ac.insert(a,k,1); 145 } 146 ac.construct(); 147 ac.work(n); 148 } 149 return 0; 150 }
hdu4511小明系列故事——女友的考验(ac自动机+最短路),布布扣,bubuko.com
hdu4511小明系列故事——女友的考验(ac自动机+最短路)
原文:http://www.cnblogs.com/shangyu/p/3750801.html