1 /*
2 *zoj2676 0--1分数规划
3 * 双向边, 加边的时候要加双容量
4 */
5 #include <stdio.h>
6 #include <string.h>
7  #define INF 0x3fffffff
8  #define EPS 1e-6
9 #define NN 104
10 #define MM 804
11
12 int x[MM], y[MM];
13 double c[MM];
14 int mark[NN];
15
16 typedef struct node {
17 int v;
18 double w;
19 struct node *nxt, *op;
20 } NODE;
21 NODE edg[MM];
22 NODE *Link[NN];
23 int h[NN];
24 int num[NN];
25
26 int M, N, idx, S, T, n; // S 表示源点，T表示汇点，n表示节点个数
27
28 void Add(int u, int v, double c1) {
29 idx++;
30 edg[idx].v = v;
31 edg[idx].w = c1;
32 edg[idx].nxt = Link[u];
33 edg[idx].op = edg + idx + 1;
34 Link[u] = edg + idx;
35 idx++;
36 edg[idx].v = u;
37 edg[idx].w = c1;// 刚开始把无向边处理成有向边了，杯具
38 edg[idx].nxt = Link[v];
39 edg[idx].op = edg + idx - 1;
40 Link[v] = edg + idx;
41 }
42
43 double Min(double a, double b) {
44 return a < b ? a : b;
45 }
46
47 double aug(int u, double flow) {
48 if (u == T) return flow;
49 double l = flow; // l表示剩余容量
50 int tmp = n - 1;
51 for (NODE *p = Link[u]; p; p = p->nxt) {
52 if (h[u] == h[p->v] + 1 && (p->w > EPS)) {
53 double f = aug(p->v, Min(l, p->w));
54 l -= f;
55 p->w -= f;
56 p->op->w += f;
57 if (l < EPS || h[S] == n) return flow - l; // gap
58 }
59 if (p->w > EPS && h[p->v] < tmp) {
60 tmp = h[p->v];
61 }
62 }
63 if (l == flow) {
64 num[h[u]]--; // gap
65 if (num[h[u]] == 0) h[S] = n; // gap，每个点的距离值最多为n - 1，这里设为n 表示断层了
66 else {
67 h[u] = tmp + 1;
68 num[h[u]]++; // gap
69 }
70 }
71 return flow - l;
72 }
73
74 void Init() {
75 idx = 0;
76 S = 1;
77 T = N;
78 n = N;
79 memset(Link, 0, sizeof (Link));
80 }
81
82 /*n表示总点的个数，包括源点和汇点*/
83 double sap() {
84 double ans = 0;
85 memset(h, 0, sizeof (h));
86 memset(num, 0, sizeof (num));
87 num[0] = n;
88 while (h[S] < n) {
89 ans += aug(S, INF);
90 }
91 return ans;
92 }
93
94 double Binary() {
95 double low, hig, mid;
96 double flow;
97 int i;
98 low = 0;
99 hig = 10000001;
100 do {
101 mid = (low + hig) / 2;
102 Init();
103 flow = 0;
104 for (i = 1; i <= M; i++) {
105 if (c[i] + EPS < mid) {// 因为求得是最小割，所以所有负边可以直接加入
106 flow += c[i] - mid;
107 } else {
108 Add(x[i], y[i], c[i] - mid); // 对正边求最大流
109 }
110 }
111 flow += sap();
112 if (flow > EPS) {
113 low = mid;
114 } else {
115 hig = mid;
116 }
117 } while (low + EPS < hig);
118 return mid;
119 }
120
121 void dfs(int u){// 所有能搜到的点，即为割[S,T]中S集合的点
122 for (NODE *p = Link[u]; p; p = p->nxt){
123 if (p->w > EPS && !mark[p->v]){
124 mark[p->v] = 1;
125 dfs(p->v);
126 }
127 }
128 }
129 void Solve(){
130 int i, first;
131 double rat = Binary();// 参数搜索，得到最小比率值
132 memset(mark, 0, sizeof(mark));
133 mark[S] = 1;
134 dfs(S);
135 // 求割边数
136 int ans = 0;
137 for (i = 1; i <= M; i++){
138 if (c[i] + EPS < rat || (mark[x[i]] + mark[y[i]]) == 1){
139 ans++;
140 }
141 }
142 printf("%d\n", ans);
143 // 输出割边
144 first = 1;
145 for (i = 1; i <= M; i++){
146 if (c[i] + EPS < rat || (mark[x[i]] + mark[y[i]]) == 1){
147 if (first){
148 printf("%d", i);
149 first = 0;
150 }else printf(" %d", i);
151 }
152 }
153 puts("");
154 }
155 int main() {
156 int i;
157 int first = 1;
158 while (scanf("%d%d", &N, &M) != EOF) {
159 if (!first) puts("");
160 first = 0;
161 for (i = 1; i <= M; i++) {
162 scanf("%d%d%lf", &x[i], &y[i], &c[i]);
163 }
164 Solve();
165 }
166 return 0;
167 }
168