poj3694 Network 【图论-Tarjan-Lca】
2017-04-11 17:39
351 查看
Network
Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 9848 Accepted: 3657
Description
A network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers are connected directly or indirectly by successive links, so data can be transformed between any two computers. The administrator finds that some links are vital to the network, because failure of any one of them can cause that data can’t be transformed between some computers. He call such a link a bridge. He is planning to add some new links one by one to eliminate all bridges.
You are to help the administrator by reporting the number of bridges in the network after each new link is added.
Input
The input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤ 200,000).
Each of the following M lines contains two integers A and B ( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B. Computers are numbered from 1 to N. It is guaranteed that any two computers are connected in the initial network.
The next line contains a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links the administrator plans to add to the network one by one.
The i-th line of the following Q lines contains two integer A and B (1 ≤ A ≠ B ≤ N), which is the i-th added new link connecting computer A and B.
The last test case is followed by a line containing two zeros.
Output
For each test case, print a line containing the test case number( beginning with 1) and Q lines, the i-th of which contains a integer indicating the number of bridges in the network after the first i new links are added. Print a blank line after the output for each test case.
Sample Input
3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0
Sample Output
Case 1:
1
0
Case 2:
2
0
AC代码:
# include <cstdio> # include <cstring> # define MAXN 500005 struct EDGE { int v; int next; }edge[MAXN]; int low[MAXN]; int dfn[MAXN]; int head[MAXN]; int pre[MAXN]; int depth[MAXN]; int bridge[MAXN]; int tot, index; int cut; int casecount; void Init() { tot = 0; memset(head, -1, sizeof(head)); } void Addedge(int u, int v) { edge[tot].v = v; edge[tot].next = head[u]; head[u] = tot ++; } void Tarjan(int u) { low[u] = dfn[u] = ++index; depth[u] = depth[pre[u]] + 1; for (int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].v; if (!dfn[v]) { pre[v] = u; Tarjan(v); if (low[v] < low[u]) { low[u] = low[v]; } if (low[v] > dfn[u]) { cut ++; bridge[v] = 1; } } else if (v != pre[u]) { if (dfn[v] < low[u]) { low[u] = dfn[v]; } } } } void Lca(int u, int v) { while (depth[u] > depth[v]) { if (bridge[u]) { cut --; bridge[u] = 0; } u = pre[u]; } while (depth[u] < depth[v]) { if (bridge[v]) { cut --; bridge[v] = 0; } v = pre[v]; } while (u != v) { if (bridge[u]) { cut --; bridge[u] = 0; } if (bridge[v]) { cut --; bridge[v] = 0; } u = pre[u]; v = pre[v]; } } void Solve(int n) { int q, a, b, i; memset(bridge, 0, sizeof(bridge)); memset(dfn, 0, sizeof(dfn)); memset(depth, 0, sizeof(depth)); for (i = 1; i <= n; i++) { pre[i] = i; } for (i = 1; i <= n; i++) { pre[i] = i; } cut = 0; index = 0; Tarjan(1); scanf("%d", &q); while (q--) { scanf("%d %d", &a, &b); Lca(a, b); printf("%d\n", cut); } printf("\n"); } int main(void) { casecount = 1; int n, m; while (~scanf("%d %d", &n, &m), n && m) { printf("Case %d:\n", casecount ++); Init(); for (int i = 0; i < m; i++) { int a, b; scanf("%d %d", &a, &b); Addedge(a, b); Addedge(b, a); } Solve(n); } return 0; }
相关文章推荐
- POJ3694-Network(Tarjan缩点+LCA)
- POJ3694 Network(求桥的数目,lca,Tarjan)
- poj3694 Network 求桥边个数[tarjan + LCA]
- poj 3694 Network(tarjan + LCA)
- 7_6_B题 Network题解[POJ3694] (LCA + 求桥 + 并查集)
- 图论 LCA离线算法 Tarjan
- D - Network - poj3694(LCA求桥)
- poj 3529 Network 双连通分连 动态求桥的数目 + tarjan离线算法求LCA
- Network(Tarjan+缩点+LCA)
- poj3694(tarjan缩点+lca)
- poj3694--Network(双连通缩点+lca)
- Tarjan+LCA POJ 3694 Network
- POJ 3694 Network(tarjan+lca)
- POJ3694 Network【连通分量+LCA】
- 【POJ3694】Network {tarjan+并查集}
- POJ 3694 Network (tarjan bcc + LCA)
- [POJ3694]Network(LCA, 割边, 桥)
- HDU 3078 - Network (LCA)【tarjan离线/DFS倍增】
- 图论 tarjan 求 LCA 祖孙询问
- POJ3694 Network 割边 LCA