题目链接

题意： 给定一个无向图，问最少加入多少条边，使得这个图成为连通图

思路：首先注意题目给出的无向图可能是非连通的，即存在孤立点。处理孤立点之后。其它就能够当作连通块来处理。事实上跟POJ3352非常像，仅仅只是存在孤立点而已。所以找出桥，缩点，然后统计度数为0(伸出两条边)的点u和度数为1(伸出一条边)的点。最后的答案为(2 * u + v + 1) / 2。

POJ3352

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <utility>
#include <algorithm>using namespace std;const int MAXN = 1005;struct Edge{int to, next;
}edge[MAXN * 100];int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], dg[MAXN], used[MAXN];
int Index;
int bridge;void addedge(int u, int v) {edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;
}void Tarjan(int u, int pre) {int v;  Low[u] = DFN[u] = ++Index;  for (int i = head[u]; i != -1; i = edge[i].next) {  v = edge[i].to;  if (v == pre) continue;   if (!DFN[v]) {  Tarjan(v, u);   if (Low[u] > Low[v]) Low[u] = Low[v];  if (Low[v] > DFN[u])bridge++;}   else if (Low[u] > DFN[v])  Low[u] = DFN[v];  }
}void init() {memset(head, -1, sizeof(head));memset(DFN, 0, sizeof(DFN)); memset(Low, 0, sizeof(Low)); Index = tot = 0;bridge = 0;
}void solve(int N) {for (int i = 1; i <= N; i++)if (!DFN[i]) {Tarjan(i, i);bridge++;}if (bridge == 1) {printf("0\n"); }else {memset(used, 0, sizeof(used));memset(dg, 0, sizeof(dg)); for (int u = 1; u <= N; u++) {if (head[u] == -1) {used[Low[u]] = 1;continue;}for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].to; used[Low[u]] = used[Low[v]] = 1;if (Low[u] != Low[v]) {dg[Low[u]]++; }}}int ans = 0;for (int u = 1; u <= N; u++) {if (used[u] && dg[u] == 0) ans += 2; else if (dg[u] == 1) ans++; }ans = (ans + 1) / 2;printf("%d\n", ans);}
}int main() {int n, m; while (scanf("%d%d", &n, &m) != EOF) {init(); int u, v;while (m--) {scanf("%d%d", &u, &v); addedge(u, v);addedge(v, u);}solve(n);}return 0;
}