AOJ GRL_1_B: Shortest Path

题目链接：

　　http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=GRL_1_B

Single Source Shortest Path (Negative Edges)

Input
An edge-weighted graph G (V, E) and the source r.

|V| |E| r
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 in G. The graph vertices are named with the numbers 0, 1,…, |V|−1 respectively. r is the source of the graph.

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

Output
If the graph contains a negative cycle (a cycle whose sum of edge costs is a negative value) which is reachable from the source r, print

NEGATIVE CYCLE
in a line.

Otherwise, print

c0
c1
:
c|V|−1
The output consists of |V| lines. Print the cost of the shortest path from the source r to each vertex 0, 1, … |V|−1 in order. If there is no path from the source to a vertex, print “INF”.

Constraints
1 ≤ |V| ≤ 1000
0 ≤ |E| ≤ 2000
-10000 ≤ di ≤ 10000
There are no parallel edges
There are no self-loops
Sample Input 1
4 5 0
0 1 2
0 2 3
1 2 -5
1 3 1
2 3 2
Sample Output 1
0
2
-3
-1

Sample Input 2
4 6 0
0 1 2
0 2 3
1 2 -5
1 3 1
2 3 2
3 1 0
Sample Output 2
NEGATIVE CYCLE

Sample Input 3
4 5 1
0 1 2
0 2 3
1 2 -5
1 3 1
2 3 2
Sample Output 3
INF
0
-5
-3

这题求单源最短路径+负圈，用Bellman-Ford算法。

注意：这里求的负圈附带了条件，就是源点r能触及到的负圈。

Bellman-Ford算法+求负圈链接

代码：

#include <iostream>#include <algorithm>#include <map>#include <vector>using namespace std;typedef long long ll;#define INF 2147483647struct edge{    int from,to,cost; };edge es[2010];  //存储边 int d[1010];  // d[i] 表示点i离源点的最短距离 int V,E; //点和边的数量 bool shortest_path(int s){    fill(d,d+V,INF);    d[s] = 0;    int v = 0;    while(true){        bool update = false;        for(int i = 0;i < E; i++){            edge e = es[i];            if(d[e.from] != INF && d[e.to] > d[e.from] + e.cost){                d[e.to] = d[e.from] + e.cost;                update = true;                if(v == V-1) return true;            }        }        if(!update) break;        v++;    }    return false;}int main(){    int r;    cin >> V >> E >> r;    for(int i = 0;i < E; i++) cin >> es[i].from >> es[i].to >>es[i].cost;    if(!shortest_path(r)){        for(int i = 0;i < V; i++){            if(d[i] == INF) cout <<"INF" <<endl;            else cout << d[i] << endl;        }    }else{        cout <<"NEGATIVE CYCLE" <<endl;    }    return 0;}