Bzoj 2834: 回家的路 dijkstra,堆优化,分层图,最短路
2834: 回家的路
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 62 Solved: 38
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Description
Input
Output
Sample Input
2 1
1 2
1 1 2 2
1 2
1 1 2 2
Sample Output
5
HINT
N<=20000,M<=100000
Source
dijkstra+堆优化+分层图
把所有的横向和纵向分开看。跑最短路即可。
注意：N这么大，不能写N^2建图。要把M个位置去建图。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define MAXN 20010 4 #define MAXM 100010 5 #define INF 1e9 6 struct NODE 7 { 8 int begin,end,value,next; 9 }edge[6*MAXM+10]; 10 struct node 11 { 12 int x,y,id; 13 }a[MAXM+10]; 14 int cnt,Head[2*MAXM+10],pos[2*MAXM+10],Heap[2*MAXM+10],dis[2*MAXM+10],N,SIZE; 15 void addedge(int bb,int ee,int vv) 16 { 17 edge[++cnt].begin=bb;edge[cnt].end=ee;edge[cnt].value=vv;edge[cnt].next=Head[bb];Head[bb]=cnt; 18 } 19 void addedge1(int bb,int ee,int vv) 20 { 21 addedge(bb,ee,vv);addedge(ee,bb,vv); 22 } 23 int read() 24 { 25 int s=0,fh=1;char ch=getchar(); 26 while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} 27 while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} 28 return s*fh; 29 } 30 //int xy(int x,int y){return (x-1)*n+y;} 31 void Push1(int k) 32 { 33 int now=k,root; 34 while(now>1) 35 { 36 root=now/2; 37 if(dis[Heap[root]]<=dis[Heap[now]])return; 38 swap(Heap[root],Heap[now]); 39 swap(pos[Heap[root]],pos[Heap[now]]); 40 now=root; 41 } 42 } 43 void Insert(int k) 44 { 45 Heap[++SIZE]=k;pos[k]=SIZE;Push1(SIZE); 46 } 47 void Pop1(int k) 48 { 49 int now,root=k; 50 pos[Heap[k]]=0;Heap[k]=Heap[SIZE--];if(SIZE>0)pos[Heap[k]]=k; 51 while(root<=SIZE/2) 52 { 53 now=root*2; 54 if(now<SIZE&&dis[Heap[now+1]]<dis[Heap[now]])now++; 55 if(dis[Heap[root]]<=dis[Heap[now]])return; 56 swap(Heap[root],Heap[now]); 57 swap(pos[Heap[root]],pos[Heap[now]]); 58 root=now; 59 } 60 } 61 void dijkstra(int start) 62 { 63 int i,u,v; 64 for(i=1;i<=N;i++)dis[i]=INF;dis[start]=0; 65 for(i=1;i<=N;i++)Insert(i); 66 while(SIZE>0) 67 { 68 u=Heap[1];Pop1(pos[u]); 69 for(i=Head[u];i!=-1;i=edge[i].next) 70 { 71 v=edge[i].end; 72 if(dis[v]>dis[u]+edge[i].value){dis[v]=dis[u]+edge[i].value;Push1(pos[v]);} 73 } 74 } 75 } 76 bool cmp1(node aa,node bb) 77 { 78 if(aa.x==bb.x)return aa.y<bb.y; 79 return aa.x<bb.x; 80 } 81 bool cmp2(node aa,node bb) 82 { 83 if(aa.y==bb.y)return aa.x<bb.x; 84 return aa.y<bb.y; 85 } 86 int main() 87 { 88 int n,m,i,k,k1,bx,by,ex,ey; 89 n=read();m=read(); 90 memset(Head,-1,sizeof(Head));cnt=1; 91 N=2*m+4; 92 for(i=1;i<=m+2;i++)a[i].x=read(),a[i].y=read(),a[i].id=i; 93 sort(a+1,a+m+3,cmp1); 94 for(i=1;i<=m+1;i++) 95 { 96 if(a[i].x==a[i+1].x)addedge1(a[i].id,a[i+1].id,2*(a[i+1].y-a[i].y)); 97 } 98 sort(a+1,a+m+3,cmp2); 99 for(i=1;i<=m+1;i++) 100 { 101 if(a[i].y==a[i+1].y)addedge1(a[i].id+m+2,a[i+1].id+m+2,2*(a[i+1].x-a[i].x)); 102 } 103 for(i=1;i<=m;i++)addedge1(i,i+m+2,1); 104 addedge1(m+1,m+1+m+2,0);addedge1(m+2,m+2+m+2,0); 105 dijkstra(m+1); 106 if(dis[m+2]!=INF)printf("%d",dis[m+2]); 107 else printf("-1"); 108 /*不看n的范围的后果。。。写了个n^2的建图。。。 109 for(i=1;i<=m;i++) 110 { 111 x=read();y=read(); 112 k=xy(x,y); 113 addedge1(k,n*n+k,1); 114 } 115 for(i=1;i<=n;i++) 116 { 117 for(j=1;j<=n;j++) 118 { 119 if(i<n){k=xy(i,j);k1=xy(i+1,j);addedge1(k,k1,2);addedge1(k+n*n,k1+n*n,2);} 120 if(j<n){k=xy(i,j);k1=xy(i,j+1);addedge1(k,k1,2);addedge1(k+n*n,k1+n*n,2);} 121 } 122 } 123 N=2*n*n; 124 bx=read();by=read();ex=read();ey=read(); 125 addedge1(xy(bx,by),xy(bx,by)+n*n,0); 126 addedge1(xy(ex,ey),xy(ex,ey)+n*n,0); 127 dijkstra(xy(bx,by)); 128 if(dis[xy(ex,ey)]!=INF)printf("%d",dis[xy(ex,ey)]); 129 else printf("-1");*/ 130 fclose(stdin); 131 fclose(stdout); 132 return 0; 133 }
