题目链接:
http://codeforces.com/gym/100114
Description
The computer network of “Plunder & Flee Inc.” consists of n servers and m two-way communication links. Two servers can communicate either through a direct link, or through a chain of links, by relaying information from server to server. Current network setup enables communication for any pair of servers. The network administrator strives to maximize network reliability. Some communication links in the network were identified as critical. A failure on any critical link will split the network into disconnected segments. Company management responded to the administrator’s concerns and agreed to fund another communication link, provided that when the new link goes online the number of critical links will be minimized. Write a program that, given a network configuration, will pick a pair of servers to be connected by the new communication link. If several such pairs allow minimizing the number of critical links then any of them will be considered as a correct answer. Example. The following figure presents a network consisting of 7 servers and 7 communication links. Essential links are shown as bold lines. A new link connecting servers #1 and #7 (dotted line) can reduce the number of the critical links to only one
Input
Output
Sample Input
7 7 1 2 2 3 2 4 2 6 3 4 4 5 6 7
Sample Output
HINT
题意:
题解:
代码:
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<stack>
using namespace std; const int maxn = + ; struct Edge {
int u,v,type;
Edge(int u,int v,int type=) : u(u),v(v),type(type) {}
}; vector<int> G[maxn];
vector<Edge> egs;
int n, m; void addEdge(int u, int v) {
egs.push_back(Edge(u,v));
G[u].push_back(egs.size() - );
} int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;
stack<int> S; //无相图的边双联通分量 tarjan
void dfs(int u) {
pre[u] = lowlink[u] = ++dfs_clock;
S.push(u);
for (int i = ; i < G[u].size(); i++){
Edge& e = egs[G[u][i]];
if (e.type == ) continue;
egs[G[u][i] ^ ].type ^= ;//针对无相图,要标记反向边
if (!pre[e.v]) {
dfs(e.v);
lowlink[u] = min(lowlink[u], lowlink[e.v]);
}
else if (!sccno[e.v]) {
lowlink[u] = min(lowlink[u],pre[e.v]);
}
}
if (lowlink[u] == pre[u]) {
scc_cnt++;
for (;;) {
int x = S.top(); S.pop();
sccno[x] = scc_cnt;
if (x == u) break;
}
}
} void find_scc() {
dfs_clock = scc_cnt = ;
memset(sccno, , sizeof(sccno));
memset(pre, , sizeof(pre));
for (int i = ; i < n; i++) if (!pre[i]) dfs(i);
} vector<int> T[maxn];
//缩点得到的桥构成的树
void build_tree() {
for (int i = ; i < egs.size(); i++) {
Edge& e = egs[i];
if (e.type) continue;
if (sccno[e.u] != sccno[e.v]) {
T[sccno[e.u]].push_back(sccno[e.v]);
T[sccno[e.v]].push_back(sccno[e.u]);
}
}
//for (int i = 1; i <= scc_cnt; i++) {
// printf("%d:", i);
// for (int j = 0; j < T[i].size(); j++) {
// int v = T[i][j];
// printf("%d ", v);
// }
// printf("\n");
//}
} //求树的直径
void dfs2(int u,int fa,int d,int &dep,int &res) {
if (dep < d) dep = d, res = u;
for (int i = ; i < T[u].size(); i++) {
int v = T[u][i];
if (v == fa) continue;
dfs2(v, u, d + , dep, res);
}
} void init() {
for (int i = ; i < n; i++) G[i].clear(), T[i].clear();
T[n].clear();
} int main() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
scanf("%d%d", &n, &m);
init();
for (int i = ; i < m; i++) {
int u, v;
scanf("%d%d", &u, &v),u--,v--;
addEdge(u, v);
addEdge(v, u);
}
find_scc();
build_tree();
int u, v, dep;
dep=-,dfs2(, -, , dep, u);
dep=-,dfs2(u, -, , dep, v);
int au, av;
for (int i = ; i < n; i++) {
//printf("sccno[%d]:%d\n", i,sccno[i]);
if (sccno[i] == u) au = i + ;
if (sccno[i] == v) av = i + ;
}
printf("%d %d\n", au, av);
return ;
}