弗洛伊德算法模板 All Pairs Shortest Path Aizu - GRL_1_C

題目：

Input

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

|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1

|V| is the number of vertices and |E| is the number of edges inG. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

s_i and t_i represent source and target vertices ofi-th edge (directed) and d_i represents the cost of thei-th edge.

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

D0,0 D0,1 ... D0,|V|-1
D1,0 D1,1 ... D1,|V|-1
:
D|V|-1,0 D1,1 ... D|V|-1,|V|-1

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

Constraints

• 1 ≤ |V| ≤ 100
• 0 ≤ |E| ≤ 9900
• -2 × 10⁷ ≤ d_i ≤ 2 × 10⁷
• There are no parallel edges
• There are no self-loops

Sample Input 1

4 6
0 1 1
0 2 5
1 2 2
1 3 4
2 3 1
3 2 7

Sample Output 1

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

Sample Input 2

4 6
0 1 1
0 2 -5
1 2 2
1 3 4
2 3 1
3 2 7

Sample Output 2

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

Sample Input 3

4 6
0 1 1
0 2 5
1 2 2
1 3 4
2 3 1
3 2 -7

Sample Output 3

NEGATIVE CYCLE

題意：

首先輸入有向帶權圖的點數v和邊數e，接下來e行，每行輸入三個數s，t，d，表示一條s到t的權爲d的邊。

若圖中存在負環,則輸出：NEGATIVE CYCLE。否則輸出所有點對的最短路（矩陣）。

#include<bits/stdc++.h>

typedef long long ll;

using namespace std;

const ll INF = 1LL<<32;

int n,e;

ll dp[110][110];

void floyd(){
	for(int k = 0 ; k<n ; k++){
		for(int i = 0 ; i<n ; i++){
			if(dp[i][k] == INF)
				continue;
			for(int j = 0 ; j<n ; j++){
				if(dp[k][j] == INF)
					continue;
				dp[i][j] = min(dp[i][j] , dp[i][k]+dp[k][j]);
			}
		}
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n>>e;
	
	for(int i = 0 ; i<n ; i++){
		for(int j = 0 ; j<n ; j++){
			if(i == j)
				dp[i][j] = 0;
			else
				dp[i][j] = INF; 
		}
	}
	
	for(int i = 0 ; i<e ; i++){
		int s,t,d;
		cin>>s>>t>>d;
		dp[s][t] = d;
	}
	
	floyd();
	
	bool flag = 0;
	for(int i = 0 ; i<n ; i++){
		if(dp[i][i] < 0){
			flag = 1;
			break;
		}
	}
	if(flag){
		cout<<"NEGATIVE CYCLE"<<endl;
	}
	else{
		for(int i = 0 ; i<n ; i++){
			for(int j = 0 ; j<n ; j++){
				if(j) cout<<" ";
				if(dp[i][j] == INF){
					cout<<"INF";
				}
				else
					cout<<dp[i][j];
			}
			cout<<endl;
		}
	}
	
	return 0;
}