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

题目：

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.

si and ti represent source and target vertices ofi-th edge (directed) and
di 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 ≤ di ≤ 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

题意：

首先输入有向带权图的点数v和边数e，接下来e行，每行输入三个数s，t，d，表示一条s到t的权为d的边。

若图中存在负环,则输出：NEGATIVE CYCLE。否则输出所有点对的最短路（矩阵）。

#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;
}