Warm up

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 10317 Accepted Submission(s): 2380

Problem Description

　　N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
　　If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
　　Note that there could be more than one channel between two planets.

Input

　　The input contains multiple cases.
　　Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
　　(2<=N<=200000, 1<=M<=1000000)
　　Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
　　A line with two integers '0' terminates the input.

Output

　　For each case, output the minimal number of bridges after building a new channel in a line.

Sample Input

4 4 1 2 1 3 1 4 2 3 0 0

Sample Output

题目大意：有一个连通的无向图，现在要给这张图添加一条边，让你求添加一条边以后，图中桥的数量的最小值

思路：如果知道树的直径这个概念的话，思路就不难想到，要想使得桥的数量最小，那么一定让图的双连通性最大，也就是一条边可以让最多的点形成双联通，也就是树的直径了，所以先缩点，注意处理重边，然后对这棵树进行两次dfs求出树的直径。

这些操作都很简单，主要是题目对时间和内存要求比较高，平时这两个地方不严谨的话很容易TLE或MLE

经过多次实验，最稳妥的办，法就是数组卡着内存开，点多用链式前向星，缩完点后点少了用vector

Accepted code

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
using namespace std;

#define sc scanf
#define ls rt << 1
#define rs ls | 1
#define Min(x, y) x = min(x, y)
#define Max(x, y) x = max(x, y)
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define MEM(x, b) memset(x, b, sizeof(x))
#define lowbit(x) ((x) & -(x))
#define P2(x) ((x) * (x))

typedef long long ll;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 100;
const int MAXM = 2e5 + 100;
const int INF = 0x3f3f3f3f;
inline ll fpow(ll a, ll b){ ll r = 1, t = a; while (b){ if (b & 1)r = (r*t) % MOD; b >>= 1; t = (t*t) % MOD; }return r; }

struct Edge
{
	int v, f, next;
}e[MAXN << 1];
vector <int> edge[MAXM];
int head[MAXM], n, m, cnt, tot, stot;
int dfn[MAXM], low[MAXM], suo[MAXM],  max_1, rt;
stack <int> st;
bool vis[MAXM];
void init() {
	MEM(head, -1); MEM(e, 0); MEM(vis, 0);
	MEM(dfn, 0); MEM(low, 0); MEM(suo, 0);
	cnt = tot = stot = 0;
}
void add(int from, int to) {
	e[cnt].v = to;
	e[cnt].next = head[from];
	head[from] = cnt++;
}
void tarjan(int x) {
	dfn[x] = low[x] = ++tot;
	st.push(x);
	for (int i = head[x]; i != -1; i = e[i].next) {
		int vi = e[i].v;
		if (e[i].f) continue;
		e[i].f = e[i ^ 1].f = 1;
		if (!dfn[vi]) {
			tarjan(vi);
			Min(low[x], low[vi]);
		}
		else Min(low[x], dfn[vi]);
	}
	if (dfn[x] == low[x]) {
		stot++; int k;
		do {
			k = st.top(); st.pop();
			suo[k] = stot;
		} while (k != x);
	}
}
void dfs(int x, int step) {
	vis[x] = true;
	if (step > max_1) max_1 = step, rt = x;
	for (int i = 0; i < SZ(edge[x]); i++) {
		int vi = edge[x][i];
		if (vis[vi]) continue;
		dfs(vi, step + 1);
	}
}

int main()
{
	while (~sc("%d %d", &n, &m) && n + m) {
		init();
		for (int i = 0; i < m; i++) {
			int ui, vi; sc("%d %d", &ui, &vi);
			add(ui, vi); add(vi, ui);
		}
		tarjan(1);
		for (int i = 1; i <= n; i++) {
			for (int j = head[i]; j != -1; j = e[j].next) {
				int ui = suo[i], vi = suo[e[j].v];
				if (ui != vi) edge[ui].push_back(vi);
			}
		}
		max_1 = 0; dfs(1, 0);
		MEM(vis, 0);
		max_1 = 0; dfs(rt, 0);
		printf("%d\n", stot - max_1 - 1);  // 树的大小 - 直径
		for (int i = 1; i <= stot; i++) edge[i].clear();
	}
	return 0;
}

#HDU 4612 Warm up (tarjan + 树的直径)

Warm up

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