AOJ grl_1_c:all Pairs Shortest path (Floyd-warshall algorithm to find the shortest path between any two points) (Bellman-ford algorithm to determine negative circle)

Title Link: Http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_C

All Pairs Shortest Path

Input

An edge-weighted graph G ( V , E ).

|V| |E|s0t0d0s1t1d1:s|E|?1t|E|?1d|E|?1

|V|Is the number of vertices and are the number of |E| edges in G . The graph vertices is named with the numbers 0, 1,..., |V|?1 respectively.

siand ti represent source and target vertices of i -th edge (directed) and represents the cost of the di i -th E Dge.

Output

If The graph contains a negative cycle (a cycle whose sum of edge costs is a negative value), print

Negative CYCLE

In a line.

Otherwise, Print

D0,0D0,1D0,|V|?1D1,0D1,1D1,|V|?1:D|V|?1,0D1,1D|V|?1,|V|?1

The output consists of |V| lines. For each i th line, print the cost of the shortest path from vertex to each i vertex j ( j=0,1,…|V|?1 ) respectively . If there is no path from vertex i to vertex j , print "INF". Print a space between the costs.

Constraints

• 1 ≤ |V| ≤100
• 0≤ |E| ≤9900
• -2x107≤ di ≤2x107
• There is no parallel edges
• There is no self-loops

Sample Input 1

4 60 1 10 2 51 2 21 3 42 3 13 2 7

Sample Output 1

0 1 3 4INF 0 2 3INF inf 0 1INF inf 7 0

Sample Input 2

4 60 1 10 2-51 2 21 3 42 3 13 2 7

Sample Output 2

0 1-5 -4inf 0 2 3INF inf 0 1INF inf 7 0

Sample Input 3

4 60 1 10 2 51 2 21 3 42 3 13 2-7

Sample Output 3

Negative CYCLE

This problem first uses the Bellman-ford algorithm to judge the negative circle, then uses the Floyd-warshall algorithm to seek any two points between the shortest circuit.

Code:

#include <iostream>#include<algorithm>#include<map>#include<vector>using namespaceStd;typedefLong Longll;#defineINF 2147483647structedge{int  from, To,cost;}; Edge es[10000];intd[ the][ the];//D[i][j] Indicates the shortest path of point I to JintV,e;//number of points and edges//Judging negative CirclesBOOLFind_negative_loop () {ints[ the]; Fill (s,s+v,0);  for(inti =0; i < V; i++){         for(intj =0; J < E; J + +) {Edge e=Es[j]; if(S[e.to] > S[e. from] +e.cost) {S[e.to]= S[e. from] +E.cost; if(i = = V1)return false; }        }    }    return true;}//Shortest path between any two pointsvoidWarshall_floyd () { for(intK =0; k < V; k++){         for(inti =0; i < V; i++){             for(intj =0; J < V; J + +){                if(D[i][k]! = INF && d[k][j]! =INF) D[i][j]= Min (D[i][j], d[i][k] +D[k][j]); }        }    }}intMain () {CIN>> V >>E;  for(inti =0; i < V; i++){         for(intj =0; J < V; J + +) {D[i][j]=INF; } D[i][i]=0; }     for(inti =0; i < E; i++) Cin >> Es[i]. from>> es[i].to >> Es[i].cost,d[es[i]. from][es[i].to] =Es[i].cost; if(Find_negative_loop ()) {Warshall_floyd ();  for(inti =0; i < V; i++){             for(intj =0; J < V; J + +){                if(J! =0) cout <<" "; if(D[i][j] = = INF) cout <<"INF"; Elsecout <<D[i][j]; } cout<<Endl; }    }Else{cout<<"Negative CYCLE"<<Endl; }        return 0;} 

AOJ grl_1_c:all Pairs Shortest path (Floyd-warshall algorithm to find the shortest path between any two points) (Bellman-ford algorithm to determine negative circle)