給定一個圖,並給定邊,a b c(c==1||c==2) 表示ab之間有c條邊 求把儘可能多的有向邊定向變成強聯通圖。
先把圖當做無向圖,加邊時記錄是否有邊,dfs的時候不要把本沒有的邊用到!因為這個錯了好多次。。。。然後就簡單了,記錄橋就可以了。
- /**************************************************
- Problem: 1438 User: G_lory
- Memory: 5312K Time: 657MS
- Language: C++ Result: Accepted
- **************************************************/
- #include <cstdio>
- #include <cstring>
- #include <iostream>
- #define pk printf("KKK!\n");
- using namespace std;
- const int N = 2005;
- const int M = N * N;
- struct Edge {
- int from, to, next;
- int flag; // 1代表單向邊 0代表沒邊 2代表雙向邊
- int cut;
- } edge[M];
- int cnt_edge;
- int head[N];
- void add_edge(int u, int v, int c)
- {
- edge[cnt_edge].from = u;
- edge[cnt_edge].to = v;
- edge[cnt_edge].next = head[u];
- edge[cnt_edge].flag = c;
- edge[cnt_edge].cut = 0;
- head[u] = cnt_edge++;
- }
- int dfn[N]; int idx;
- int low[N];
- int n, m;
- void tarjan(int u, int pre)
- {
- dfn[u] = low[u] = ++idx;
- for (int i = head[u]; i != -1; i = edge[i].next)
- {
- int v = edge[i].to;
- if (edge[i].flag == 0) continue;
- if (edge[i].cut == 0)
- {
- edge[i].cut = 1;
- edge[i ^ 1].cut = -1;
- }
- if (v == pre) continue;
- if (!dfn[v])
- {
- tarjan(v, u);
- low[u] = min(low[u], low[v]);
- if (dfn[u] < low[v])
- {
- edge[i].cut = 2;
- edge[i ^ 1].cut = -1;
- }
- }
- else low[u] = min(low[u], dfn[v]);
- }
- }
- void init()
- {
- idx = cnt_edge = 0;
- memset(dfn, 0, sizeof dfn);
- memset(head, -1, sizeof head);
- }
- void solve()
- {
- for (int i = 0; i < cnt_edge; ++i)
- if (edge[i].flag == 2 && (edge[i].cut == 1 || edge[i].cut == 2))
- printf("%d %d %d\n", edge[i].from, edge[i].to, edge[i].cut);
- }
- int main()
- {
- //freopen("in.txt", "r", stdin);
- while (~scanf("%d%d", &n, &m))
- {
- if (n == 0 && m == 0) break;
- int u, v, c;
- init();
- for (int i = 0; i < m; ++i)
- {
- scanf("%d%d%d", &u, &v, &c);
- add_edge(u, v, c);
- if (c == 1) c = 0;
- add_edge(v, u, c);
- }
- tarjan(1, -1);
- solve();
- }
- return 0;
- }