题意:
N≤2×105个点,M≤106条边的无向图,有重边
现在要添加一条边,问添加后剩余最小的桥数是多少
分析:
套路题,边双连通缩点建树之后,显然要形成一个最大的环让桥数变的最少
显然连树的直径的2个叶子,然后就是答案了
时间复杂度为O(n+m)
代码:
//
// Created by TaoSama on 2016-02-28
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 2e6 + 10;
int n, m;
struct Edge {
int v, nxt;
} edge[M], raw[M];
int head[N], rawHead[N], cnt;
void addRawEdge(int u, int v) {
raw[cnt] = (Edge) {v, rawHead[u]};
rawHead[u] = cnt++;
raw[cnt] = (Edge) {u, rawHead[v]};
rawHead[v] = cnt++;
}
void addEdge(int u, int v) {
edge[cnt] = (Edge) {v, head[u]};
head[u] = cnt++;
}
int dfn[N], low[N], in[N], id[N], bcc, dfsNum;
int stk[N], top;
void tarjan(int u, int f) {
dfn[u] = low[u] = ++dfsNum;
stk[++top] = u;
in[u] = true;
for(int i = rawHead[u]; ~i; i = raw[i].nxt) {
int v = raw[i].v;
if(i == (f ^ 1)) continue;
if(!dfn[v]) {
tarjan(v, i);
low[u] = min(low[u], low[v]);
} else if(in[v]) low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u]) {
++bcc;
while(true) {
int v = stk[top--];
in[v] = false;
id[v] = bcc;
if(v == u) break;
}
}
}
void init() {
bcc = dfsNum = 0;
memset(dfn, 0, sizeof dfn);
}
pair<int, int> diameter;
void dfs(int u, int f, int d) {
diameter = max(diameter, {d, u});
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(v == f) continue;
dfs(v, u, d + 1);
}
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &m) == 2 && (n || m)) {
cnt = 0; memset(rawHead, -1, sizeof rawHead);
while(m--) {
int u, v; scanf("%d%d", &u, &v);
addRawEdge(u, v);
}
init();
tarjan(1, -1);
cnt = 0; memset(head, -1, sizeof head);
int treeEdge = 0;
for(int i = 1; i <= n; ++i) {
int u = id[i];
for(int j = rawHead[i]; ~j; j = raw[j].nxt) {
int v = id[raw[j].v];
if(u == v) continue;
++treeEdge;
addEdge(u, v);
}
}
treeEdge >>= 1;
diameter = { -1, -1};
dfs(1, -1, 0);
dfs(diameter.second, -1, 0);
printf("%d\n", treeEdge - diameter.first);
}
return 0;
}