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Problem H. Hard Test
Time Limit: 20 Sec
Memory Limit: 256 MB
題目串連
http://codeforces.com/gym/100342/attachments
Description
Andrew is having a hard time preparing his 239-th contest for Petrozavodsk. This time the solution to the problem is based on Dijkstra algorithm and Andrew wants to prepare the hard test for the algorithm.
The Dijkstra algorithm is used to find the shortest path from a source vertex to all other vertices in a graph. The algorithm acts as follows. Let G be a weight directed graph with vertex set V , edge set E and weight function w : E → R +. Let all vertices be reachable from vertex s. The algorithm uses a set
of vertices U, first initialized as empty. Each vertex is labeled with either an integer number, or with +∞. Initially all vertices are labeled with +∞, and the vertex s is labeled with 0. Denote the label of vertex v as d[v].
A step of the algorithm is the following: the vertex with the minimal label that doesn’t belong to U is selected. Let this vertex be u. The vertex u is added to the set U, and each edge uv ∈ E is relaxed. The relaxation replaces d[v] with min(d[v], d[u] + w(uv)). The algorithm is over when all vertices belong to U. If the label of the vertex v has changed, the relaxation is said to be active.
Now Andrew would like to create a graph with n vertices and m edges, such that the Dijkstra algorithm makes as many active relaxations as possible. Help him to create such graph. To avoid nondeterminism, each time when selecting a vertex with minimal label among vertices that are not in U there must be exactly one vertex with the minimal label.
Input
The first line of the input file contains two integer numbers: n and m — the number of vertices and the number of edges in the graph Andrew would like to create (4 ≤ n ≤ 1000, n − 1 ≤ m ≤ n 2/5).
Output
Output m lines — the edges of the graph. Each line must contain three integer numbers: the beginning of the edge, the end of the edge and the weight of the edge. All weights must be non-negative and must not exceed 106 . All vertices must be reachable from vertex 1. If Dijkstra algorithm is run with s = 1 there must be maximal possible number of active relaxations among all graphs with n vertices and m edges. There must be no loops and no parallel edges.
Sample Input
4 3
Sample Output
1 2 0
1 3 1
1 4 2
HINT
題意
讓你出數據卡用堆最佳化的迪傑斯特拉演算法,要求鬆弛操作最多
題解:
最多的情況就是每條邊都會使得邊鬆弛一次,然後構造的話,先構造一條邊長為0的鏈,然後再從後面不斷插入就好了= =
其實我感覺我說的不是很清楚,看代碼吧……
注意,得插入有向邊
代碼:
#include <cstdio>#include <cmath>#include <cstring>#include <ctime>#include <iostream>#include <algorithm>#include <set>#include <vector>#include <sstream>#include <queue>#include <typeinfo>#include <fstream>#include <map>#include <stack>typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define test freopen("test.txt","r",stdin)#define maxn 2001#define mod 1000000007#define eps 1e-9const int inf=0x3f3f3f3f;const ll infll = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f;}//**************************************************************************************struct node{ int x,y,z;};vector<node> Q;int main(){ freopen("test.in","r",stdin); freopen("test.out","w",stdout); int n=read(),m=read(); for(int i=2;i<=n;i++) { Q.push_back((node){i-1,i,0}); m--; } for(int i=n;i>=2;i--) { if(m==0) break; for(int j=i-2;j>=1;j--) { if(m==0) break; Q.push_back((node){i,j,i-j}); m--; if(m==0) break; } if(m==0) break; } for(int i=0;i<Q.size();i++) { if(Q[i].x>Q[i].y) swap(Q[i].x,Q[i].y); printf("%d %d %d\n",Q[i].x,Q[i].y,Q[i].z); } return 0;}
Codeforces Gym 100342H Problem H. Hard Test 構造題,卡迪傑斯特拉