分层图最短路（ LYOi Online Judge 初中的最后一天）

代码参照：   LYOI Online Judge     #374. 初中的最后一天

分层图最短路模板题

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<queue>
  6
  7 using namespace std;
  8
  9 const int MAXN=5010;
 10 const int MAXM=125100;
 11 const int MOD = 1e9+7;
 12 int n,m,k,cnt;
 13 int head[MAXN];
 14 int d[MAXN][60];
 15
 16 struct Edge{
 17     int s,t,w,next;
 18 }edge[MAXM<<1];
 19
 20 inline int read(){
 21     int s=0,w=1;
 22     char ch=getchar();
 23     while (ch<'0'||ch>'9'){
 24         if(ch == '-')     w  = -1;
 25             ch = getchar();
 26     }
 27     while (ch>='0'&&ch<='9'){
 28         s=s*10+ch-'0';
 29         ch=getchar();
 30     }
 31     return s*w  ;
 32 }
 33
 34 void Add(int u,int v ,int wi ){
 35     cnt++ ;
 36     edge[cnt].s=u;
 37     edge[cnt].t=v;
 38     edge[cnt].w=wi;
 39     edge[cnt].next = head[u];
 40     head[u] = cnt;
 41 }
 42
 43 struct HeapNode{
 44     int pos , dis , v ;
 45     bool operator < ( const HeapNode  &a ) const {
 46         return  a.dis < dis ;
 47     }
 48 };
 49
 50 void dijkstra( ){
 51     priority_queue<HeapNode>Q;
 52     for(int i = 1;i <= n;i++){
 53         for(int j = 0 ; j <= k; j++ ){
 54             d[i][j]=2147483647;
 55         }
 56     }
 57     d[1][0] = 0 ;
 58     Q.push((HeapNode){1,0,0});
 59     while(!Q.empty()){
 60         HeapNode tmp = Q.top() ;
 61         int u = tmp.pos;
 62         int t = tmp.v;
 63         if( d[u][t] != tmp.dis     ){
 64             Q.pop();
 65             continue;
 66         }
 67         Q.pop();
 68     for(int i  = head[u] ; i ; i = edge[i].next ){
 69         int to  = edge[i].t ;
 70         int wi =  edge[i].w ;
 71         if( d[to][t] > d[u][t] + wi  ){
 72             d[to][t] = d[u][t] + wi ;
 73             Q.push ( (HeapNode) {  to, d[to][t] , t  });
 74             }
 75              if( t<k &&d[to][t+1] >(d[u][t]) + (edge[i].w>>1)){
 76                d[to][t+1] = d[u][t] + (edge[i].w>>1);
 77                Q.push( (HeapNode){to , d[to][t+1], t+1 });
 78             }
 79          }
 80     }
 81 }
 82
 83 int main(){
 84
 85     n=read(); m=read();  k=read();
 86
 87     for(int i = 1 ;i <= m ; i++ ){
 88         int x , y , z ;
 89         x=read() ; y=read() ; z=read() ;
 90         Add(x,y,z);  Add(y,x,z);
 91     }
 92
 93     int time;
 94     time= read();
 95     int minn=2147483647;
 96
 97     dijkstra() ;
 98     for(int i=0;i<=k;i++){
 99         minn = min(minn , d[n][i]) ;
100     }
101
102     if(minn > time ){
103         printf("QaQ\n");
104     }
105     else printf("YES\n");
106     printf("%d",minn%MOD);
107     return 0;
108 }