POJ1062

 

#include<iostream> #include<vector> #include<queue> using namespace std; struct Node { Node(){} Node(int num,int dis):num(num),dis(dis){} int dis,num; }; struct Node2 { Node2(){} Node2(int num,int dis,int maxlevel,int minlevel):num(num),dis(dis),maxlevel(maxlevel),minlevel(minlevel){} int dis,num,maxlevel,minlevel; }; bool operator<(const Node2& n1,const Node2& n2) { return n1.dis>n2.dis; } int m; int Dijkstra(const vector<vector<Node> >& g,vector<int> levels) { priority_queue<Node2> pq; pq.push(Node2(1,0,levels[1],levels[1])); while(1) { Node2 top=pq.top(); if(top.num==0) return top.dis; for(int i=0;i!=g[top.num].size();i++) if( g[top.num][i].num==0 || top.maxlevel-levels[g[top.num][i].num]<=m &&levels[g[top.num][i].num]-top.minlevel<=m ) { pq.push(Node2(g[top.num][i].num,top.dis+g[top.num][i].dis,max(top.maxlevel,levels[g[top.num][i].num]),min(top.minlevel,levels[g[top.num][i].num]))); } pq.pop(); } } int main() { //freopen("in.txt","r",stdin); int n,p,x,num,dis; cin>>m>>n; vector<vector<Node> > g(n+1); vector<int> levels(n+1); for(int i=1;i<=n;i++) { cin>>p>>levels[i]>>x; g[i].push_back(Node(0,p)); for(int j=0; j!=x; j++) { cin>>num>>dis; g[i].push_back(Node(num,dis)); } } cout<<Dijkstra(g,levels)<<endl; } 

挺不错的一道题，需要自己建图,并且需要将dijkstra改一下。

在讨论区里见到了下面的一些数据，都过了，包括最后一组据说比较变态的数据：

1 5
10000 3 2
2 5000
3 5000
3000 4 1
4 400
3000 2 1
4 500
1000 3 1
5 100
100 2 0

5700



1 5
10000 3 2
2 5000
3 5000
3000 2 1
4 500
3000 4 1
4 400
1000 3 1
5 100
100 2 0

5700



1 5
10000 3 1
2 1000
1000 2 2
3 400
4 500
1000 1 1
5 50
1000 2 1
5 100
50 2 0

1650



1 9
100 1 8
2 80
3 80
4 80
5 80
6 80
7 80
8 80
9 80
100 1 8
1 80
3 80
4 80
5 80
6 80
7 80
8 80
9 80
100 1 8
1 80
2 80
4 80
5 80
6 80
7 80
8 80
9 80
100 1 8
1 80
2 80
3 80
5 80
6 80
7 80
8 80
9 80
100 1 8
1 80
2 80
3 80
4 80
6 80
7 80
8 80
9 80
100 1 8
1 80
2 80
3 80
4 80
5 80
7 80
8 80
9 80
100 1 8
1 80
2 80
3 80
4 80
5 80
6 80
8 80
9 80
100 1 8
1 80
2 80
3 80
4 80
5 80
6 80
7 80
9 80
100 1 8
1 80
2 80
3 80
4 80
5 80
6 80
7 80
8 80

100




4 5
800 3 2
2 1
3 2
20 1 1
4 1
30 5 1
4 2
100 4 1
5 0
5 7 0

9