题意：
设从1到每个点的最短距离为d，求删除几条边后仍然使1到每个点的距离为d，使得剩下的边最多为k
解析：
先求来一遍spfa，然后bfs遍历每条路，如果d[v] == d[u] + Node[i].w 则说明这条路要保留
注意是按着走的路的顺序输出的 wa1
注意最大值设为0x3f wa3 学到了。。。emm 用memset设置数组为0x3f是无穷大
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #include <limits.h> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d\n", a) #define plld(a) printf("%lld\n", a) #define pc(a) printf("%c\n", a) #define ps(a) printf("%s\n", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 2222222, INF = 0x7fffffff;int head[maxn], vis[maxn], nex[maxn]; LL d[maxn]; int cnt, n, m, s, t, k; vector<int> g;struct node {int u, v;LL w; }Node[maxn];void add_(int u, int v, int w) {Node[cnt].u = u;Node[cnt].v = v;Node[cnt].w = w;nex[cnt] = head[u];head[u] = cnt++; }void add(int u, int v, int w) {add_(u, v, w);add_(v, u, w); }int spfa() {mem(d, 0x3f);deque<int> Q;Q.push_front(s);d[s] = 0;vis[s] = 1;while(!Q.empty()){int u = Q.front(); Q.pop_front();vis[u] = 0;for(int i = head[u]; i != -1; i = nex[i]){int v = Node[i].v;if(d[v] > d[u] + Node[i].w){d[v] = d[u] + Node[i].w;if(!vis[v]){if(Q.empty()) Q.push_front(v);else{if(d[v] < d[Q.front()]) Q.push_front(v);else Q.push_back(v);}vis[v] = 1;}}}} }void init() {mem(head, -1);g.clear();cnt = 0; }void bfs() {queue<int> Q;Q.push(1);mem(vis, 0);vis[1] = 1;int cnt1 = 0;while(!Q.empty()){int u = Q.front(); Q.pop();for(int i = head[u]; i != -1; i = nex[i]){int v = Node[i].v;if(d[v] == d[u] + Node[i].w){if(!vis[v]){g.push_back((i + 2) / 2);vis[v] = 1;Q.push(v);if(++cnt1 == k) return;}}}} }int main() {// cout << 0x3f <<endl; init();int ans = 0;int u, v, w;cin >> n >> m >> k;for(int i = 0; i < m; i++){cin >> u >> v >> w;add(u, v, w);}if(k == 0)return puts("0");s = 1;spfa();bfs();cout << g.size() << endl;for(int i = 0; i < g.size(); i++)cout << g[i] << " ";cout << endl;return 0; }
