k短路(A*)
http://poj.org/problem?id=2449
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <queue> 5 #include <algorithm> 6 using namespace std; 7 const int maxn=1e3+10; 8 9 /** 10 求第K短的算法基于BFS搜索，当终点出队K次时，所走的总距离就是第K短路， 11 不过这样那些不该走的路会被反复的走，造成许多空间时间浪费，这时候就要用到启发式的A*搜索 12 **/ 13 14 ///from s to t 15 ///opp:from t to s 16 17 struct node 18 { 19 int d,len; 20 node *next; 21 }*e[maxn],*oppe[maxn]; 22 23 ///h:from this point to destination(predict must:predict<actual) 24 ///g:from source to this point(actual) 25 int h[maxn],unreach; 26 27 struct rec 28 { 29 int d,dist; 30 bool operator<(const rec &b) const 31 { 32 return b.dist<dist; 33 } 34 }; 35 priority_queue<rec>st; 36 37 struct noa 38 { 39 int d,dist,pre; 40 bool operator<(const noa &b) const 41 { 42 if (b.pre==pre) 43 return b.dist<dist; 44 return b.pre<pre; 45 } 46 }; 47 priority_queue<noa>ast; 48 49 bool vis[maxn]; 50 int s,t,ind,ci[maxn]; 51 52 int astar() 53 { 54 int d,dist; 55 node *p; 56 ///多组数据时需要 57 // while (!ast.empty()) 58 // ast.pop(); 59 if (h[s]!=unreach) 60 ast.push({s,0,h[s]}); 61 while (!ast.empty()) 62 { 63 d=ast.top().d; 64 dist=ast.top().dist; 65 ast.pop();///! 66 ci[d]++; 67 if (ci[d]==ind && d==t) 68 return dist;///而不是ast.top().dist 69 ///优化 70 if (ci[d]>ind) 71 continue; 72 73 p=e[d]; 74 while (p) 75 { 76 if (h[p->d]!=unreach) 77 ///aster 78 ast.push({p->d,dist+p->len,dist+p->len+h[p->d]}); 79 ///just dijkstra 80 // ast.push({p->d,dist+p->len,dist+p->len}); 81 p=p->next; 82 } 83 } 84 return -1; 85 } 86 87 void opp_dijkstra() 88 { 89 int d,dd; 90 node *p; 91 memset(h,0x7f,sizeof(h)); 92 unreach=h[0]; 93 h[t]=0; 94 st.push({t,0}); 95 while (1) 96 { 97 while (!st.empty() && vis[st.top().d]) 98 st.pop(); 99 if (st.empty()) 100 break; 101 d=st.top().d; 102 st.pop(); 103 vis[d]=1; 104 p=oppe[d];/// 105 while (p) 106 { 107 dd=p->d; 108 if (h[dd]>h[d]+p->len) 109 { 110 h[dd]=h[d]+p->len; 111 st.push({dd,h[dd]}); 112 } 113 p=p->next; 114 } 115 } 116 } 117 118 int main() 119 { 120 int a,b,c,n,m,i; 121 node *p; 122 scanf("%d%d",&n,&m); 123 for (i=1;i<=m;i++) 124 { 125 scanf("%d%d%d",&a,&b,&c); 126 p=new node(); 127 p->d=b; 128 p->len=c; 129 p->next=e[a]; 130 e[a]=p; 131 132 p=new node(); 133 p->d=a; 134 p->len=c; 135 p->next=oppe[b]; 136 oppe[b]=p; 137 } 138 scanf("%d%d%d",&s,&t,&ind); 139 ///!!! 140 if (s==t) 141 ind++; 142 // ///shortest path 143 // scanf("%d%d",&s,&t); 144 // ind=1; 145 opp_dijkstra(); 146 printf("%d",astar()); 147 return 0; 148 } 149 /* 150 2 2 151 1 2 1 152 2 1 1 153 1 1 1 154 155 3 3 156 1 2 1 157 2 3 2 158 3 2 1 159 1 1 1 160 161 3 3 162 1 2 1 163 2 3 2 164 3 1 3 165 1 1 2 166 167 3 1 168 1 2 3 169 1 3 1 170 171 3 2 172 1 2 10 173 2 3 2 174 1 3 1 175 176 3 2 177 1 2 10 178 2 3 2 179 1 3 2 180 181 4 4 182 1 3 2 183 3 2 5 184 1 4 3 185 4 2 1 186 1 2 2 187 188 x 189 4 4 190 1 2 100000 191 2 3 1 192 3 2 1 193 3 4 100000 194 1 4 1000 195 196 3 4 197 1 2 1 198 2 1 2 199 2 3 1 200 3 2 1 201 2 3 5 202 203 3 3 204 1 2 1000 205 1 3 1 206 3 2 1 207 1 2 1 208 209 */
https://www.luogu.org/problemnew/show/P2483
68分……
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <queue> 5 #include <algorithm> 6 using namespace std; 7 const int maxn=5e3+10; 8 9 struct node 10 { 11 int d; 12 double len; 13 node *next; 14 }*e[maxn],*oppe[maxn]; 15 16 ///h:from this point to destination(predict must:predict<actual) 17 ///g:from source to this point(actual) 18 double h[maxn],unreach; 19 double tot; 20 21 struct rec 22 { 23 int d; 24 double dist; 25 bool operator<(const rec &b) const 26 { 27 return b.dist<dist; 28 } 29 }; 30 priority_queue<rec>st; 31 32 struct noa 33 { 34 int d,g; 35 double dist,pre; 36 bool operator<(const noa &b) const 37 { 38 if (b.pre==pre) 39 return b.dist<dist; 40 return b.pre<pre; 41 } 42 }; 43 priority_queue<noa>ast; 44 45 bool vis[maxn]; 46 int s,t,ind,re,n,m; 47 int ci[maxn]; 48 49 void aster() 50 { 51 int d,dd,g; 52 double dist,ddist; 53 node *p; 54 ind=tot/h[1];///限制次数 55 56 if (h[s]!=unreach) 57 ast.push({s,0,0,h[s]}); 58 while (!ast.empty()) 59 { 60 d=ast.top().d; 61 dist=ast.top().dist; 62 g=ast.top().g; 63 ast.pop();///! 64 ci[d]++; 65 if (d==t) 66 { 67 if (dist>tot) 68 return; 69 tot-=dist; 70 re++; 71 ind=re+tot/dist;///限制次数，优化 72 continue; ///这题的坑爹地方：到达终点，就不能再走了 73 } 74 75 p=e[d]; 76 while (p) 77 { 78 dd=p->d; 79 ddist=dist+p->len; 80 if (h[dd]!=unreach && g!=m && ci[dd]<ind && ddist<=tot) 81 ast.push({dd,g+1,ddist,ddist+h[p->d]}); 82 p=p->next; 83 } 84 } 85 } 86 87 void opp_dijkstra() 88 { 89 int d,dd,i; 90 node *p; 91 unreach=1.0e11; 92 for (i=1;i<=n;i++) 93 h[i]=unreach; 94 95 h[t]=0; 96 st.push({t,0}); 97 while (1) 98 { 99 while (!st.empty() && vis[st.top().d]) 100 st.pop(); 101 if (st.empty()) 102 break; 103 d=st.top().d; 104 st.pop(); 105 vis[d]=1; 106 p=oppe[d];/// 107 while (p) 108 { 109 dd=p->d; 110 if (h[dd]>h[d]+p->len) 111 { 112 h[dd]=h[d]+p->len; 113 st.push({dd,h[dd]}); 114 } 115 p=p->next; 116 } 117 } 118 } 119 120 int main() 121 { 122 int a,b,i; 123 double c; 124 node *p,*pp; 125 scanf("%d%d",&n,&m); 126 scanf("%lf",&tot); 127 for (i=1;i<=m;i++) 128 { 129 scanf("%d%d%lf",&a,&b,&c); 130 p=new node(); 131 p->d=b; 132 p->len=c; 133 p->next=e[a]; 134 e[a]=p; 135 136 p=new node(); 137 p->d=a; 138 p->len=c; 139 p->next=oppe[b]; 140 oppe[b]=p; 141 } 142 s=1,t=n; 143 opp_dijkstra(); 144 ///lower the memory 145 for (i=1;i<=n;i++) 146 { 147 p=oppe[i]; 148 while (p) 149 { 150 pp=p; 151 p=p->next; 152 free(pp); 153 } 154 } 155 aster(); 156 printf("%d",re); 157 return 0; 158 }
