题意是求出图的一个边割集。。。是图的边割集不是网络流的边割集。。。使得边权平均值最小。。。解法就是01分数规划。。。amber的论上有讲的。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 405
#define maxm 800005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head
struct Edge
{
int v, next;
double c;
Edge(int v = 0, double c = 0, int next = 0) : v(v), c(c), next(next) {}
}E[maxm];
queue<int> q;
int H[maxn], cntE;
int cur[maxn];
int pre[maxn];
int cnt[maxn];
int dis[maxn];
int s, t, nv;
double flow;
void addedges(int u, int v, double c)
{
E[cntE] = Edge(v, c, H[u]);
H[u] = cntE++;
E[cntE] = Edge(u, 0, H[v]);
H[v] = cntE++;
}
void bfs()
{
memset(cnt, 0, sizeof cnt);
memset(dis, -1, sizeof dis);
q.push(t);
dis[t] = 0, cnt[0] = 1;
while(!q.empty()) {
int u = q.front();
q.pop();
for(int e = H[u]; ~e; e = E[e].next) {
int v = E[e].v;
if(dis[v] == -1) {
dis[v] = dis[u] + 1;
cnt[dis[v]]++;
q.push(v);
}
}
}
}
double isap()
{
memcpy(cur, H, sizeof H);
bfs();
flow = 0;
int u = pre[s] = s, minv, pos, e;
double f;
while(dis[s] < nv) {
if(u == t) {
f = INF;
for(int i = s; i != t; i = E[cur[i]].v) if(E[cur[i]].c < f) {
f = E[cur[i]].c;
pos = i;
}
for(int i = s; i != t; i = E[cur[i]].v) {
E[cur[i]].c -= f;
E[cur[i] ^ 1].c += f;
}
flow += f;
u = pos;
}
for(e = cur[u]; ~e; e = E[e].next) if(E[e].c > eps && dis[u] == dis[E[e].v] + 1) break;
if(~e) {
cur[u] = e;
pre[E[e].v] = u;
u = E[e].v;
}
else {
if(--cnt[dis[u]] == 0) break;
for(minv = nv, e = H[u]; ~e; e = E[e].next) if(E[e].c > eps && minv > dis[E[e].v]) {
minv = dis[E[e].v];
cur[u] = e;
}
dis[u] = minv + 1;
cnt[dis[u]]++;
u = pre[u];
}
}
return flow;
}
void init()
{
cntE = 0;
memset(H, -1, sizeof H);
}
vector<int> vec;
int a[maxn];
int b[maxn];
int color[maxn];
double c[maxn];
int n, m;
bool check(double x)
{
double ans = 0;
init();
s = 0, t = n + 1, nv = t + 1;
for(int i = 1; i <= m; i++) {
double t = c[i] - x;
if(t > 0) {
addedges(a[i], b[i], t);
addedges(b[i], a[i], t);
}
else ans += t;
}
addedges(s, 1, INF);
addedges(n, t, INF);
ans += isap();
return ans > 0;
}
void dfs(int u)
{
if(color[u]) return;
color[u] = 1;
for(int e = H[u]; ~e; e = E[e].next) if(E[e].c > eps) {
int v = E[e].v;
dfs(v);
}
}
void work()
{
for(int i = 1; i <= m; i++) scanf("%d%d%lf", &a[i], &b[i], &c[i]);
double l = 0, r = INF, res = 0;
while(r - l > eps) {
double mid = (l + r) / 2;
if(check(mid)) res = l = mid;
else r = mid;
}
check(res);
memset(color, 0, sizeof color);
dfs(s);
vec.clear();
for(int i = 1; i <= m; i++) {
double t = c[i] - res;
if(t < eps) vec.push_back(i);
else if(color[a[i]] ^ color[b[i]]) vec.push_back(i);
}
printf("%d\n", (int)vec.size());
for(int i = 0; i < vec.size(); i++) printf("%d%c", vec[i], i == vec.size() - 1 ? '\n' : ' ');
}
int main()
{
while(scanf("%d%d", &n, &m) != EOF) {
work();
}
return 0;
}