题意：给你一幅图，问有多少条路径使得去掉该条路后最短路发生变化。
思路：先起始两点求两遍单源最短路，利用s[u] + t[v] + G[u][v] = dis 找出所有最短路径，构造新图。在新图中找到所有的桥输出就可以了。
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define eps 1e-8
#define INF 0x3f3f3f3f
#define MAXN 20005
#define MAXM 100005
using namespace std; struct Edge
{
int from, to, dist, pos;
Edge(int from, int to, int dist, int pos) :from(from), to(to), dist(dist), pos(pos){};
};
struct HeapNode
{
int d, u;
HeapNode(int d, int u) :d(d), u(u){};
bool operator <(const HeapNode& rhs) const{
return d > rhs.d;
}
};
struct Dijstra
{
int n, m;
vector<Edge> edges;
vector<int> G[MAXN];
bool done[MAXN];
int d[MAXN];
int p[MAXN]; void init(int n){
this->n = n;
for (int i = ; i <= n; i++){
G[i].clear();
}
edges.clear();
} void AddEdge(int from, int to, int dist, int pos = ){
edges.push_back(Edge(from, to, dist, pos));
m = edges.size();
G[from].push_back(m - );
} void dijstra(int s){
priority_queue<HeapNode> Q;
for (int i = ; i <= n; i++){
d[i] = INF;
}
d[s] = ;
memset(done, , sizeof(done));
Q.push(HeapNode(, s));
while (!Q.empty()){
HeapNode x = Q.top();
Q.pop();
int u = x.u;
if (done[u]) continue;
done[u] = true;
for (int i = ; i < G[u].size(); i++){
Edge& e = edges[G[u][i]];
if (d[e.to] > d[u] + e.dist){
d[e.to] = d[u] + e.dist;
p[e.to] = G[u][i];
Q.push(HeapNode(d[e.to], e.to));
}
else if (d[e.to] == d[u] + e.dist){ }
}
}
}
};
int pre[MAXN], isbridge[MAXM], low[MAXN];
vector<Edge> G[MAXN];
int dfs_clock;
int dfs(int u, int father){
int lowu = pre[u] = ++dfs_clock;
//int child = 0;
for (int i = ; i < G[u].size(); i++){
int v = G[u][i].to;
if (!pre[v]){
//child++;
int lowv = dfs(v, G[u][i].pos);
lowu = min(lowu, lowv);
if (lowv > pre[u]){
isbridge[G[u][i].pos] = true;
}
}
else if (pre[v] < pre[u] && G[u][i].pos != father){
lowu = min(lowu, pre[v]);
}
}
low[u] = lowu;
return lowu;
}
Dijstra s, t;
vector<Edge> edges; int res[MAXM];
int main()
{
#ifdef ONLINE_JUDGE
freopen("important.in", "r", stdin);
freopen("important.out", "w", stdout);
#endif // OPEN_FILE
int n, m;
while (~scanf("%d%d", &n, &m)){
s.init(n);
t.init(n);
edges.clear();
int x, y, z;
for (int i = ; i <= m; i++){
scanf("%d%d%d", &x, &y, &z);
edges.push_back(Edge(x, y, z, i));
edges.push_back(Edge(y, x, z, i));
s.AddEdge(x, y, z);
s.AddEdge(y, x, z);
t.AddEdge(x, y, z);
t.AddEdge(y, x, z);
}
s.dijstra();
t.dijstra(n);
LL dis = s.d[n];
//把所有最短路径找出来，在里面找出所有的桥就是答案
for (int i = ; i < edges.size(); i++){
Edge e = edges[i];
if (s.d[e.from] + e.dist + t.d[e.to] == dis){
G[e.from].push_back(Edge(e.from, e.to, e.dist, e.pos));
G[e.to].push_back(Edge(e.to, e.from, e.dist, e.pos)); }
}
dfs_clock = ;
memset(isbridge, , sizeof(isbridge));
memset(pre, , sizeof(pre));
dfs(, -);
int ans = ;
for (int i = ; i <= m; i++){
if (isbridge[i]){
ans++;
res[ans] = i;
}
}
printf("%d\n", ans);
for (int i = ; i <= ans; i++){
printf("%d ", res[i]);
}
printf("\n");
}
}