POJ-3177（无向图添加边变双连通）

最新推荐文章于 2021-02-18 10:57:34 发布

原创最新推荐文章于 2021-02-18 10:57:34 发布 · 496 阅读

0 ·

CC 4.0 BY-SA版权

每天A一道题同时被 2 个专栏收录

145 篇文章

订阅专栏

图论

28 篇文章

订阅专栏

本文深入探讨了解决POJ 3177问题的方法，涉及图论基础、桥检测算法及代码实现，旨在帮助读者理解和解决相关问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：http://poj.org/problem?id=3177

本题几乎与3352一样，只不过3352中条件更强，两点之间不会有多条边相连，而本题中没有这个条件，所以只要加上这个判断即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAX_VERTEX  5001
#define MAX_EDGE    10001

int N, M, components;
int uu[MAX_EDGE], vv[MAX_EDGE];
struct Edge{
    int to, next;
} edge[MAX_EDGE * 2];
bool isBridge[MAX_EDGE * 2];
int first[MAX_VERTEX];
int index;
inline void addEdge(int x, int y)
{
    edge[index].to = y;
    edge[index].next = first[x];
    first[x] = index++;
}

int dfsClock, low[MAX_VERTEX];
void tarjan(int x, int fa)
{
    int y, lowy, lowx = low[x] = ++dfsClock;
    bool back = true;
    for(int i = first[x]; i != -1; i = edge[i].next){
        int y = edge[i].to;
        if(y == fa && back){
            back = false;
            continue;
        }
        if(!low[y]) tarjan(y, x);
        if(low[y] > low[x]){
            isBridge[i] = isBridge[i ^ 1] = true;
        }
        lowx = min(lowx, low[y]);
    }
    low[x] = lowx;
}
void shrink(int x)
{
    low[x] = components;
    for(int i = first[x]; i != -1; i = edge[i].next){
        if(!isBridge[i] && !low[edge[i].to])
            shrink(edge[i].to);
    }
}
void solve()
{
//mark all bridges
    memset(low + 1, 0, N << 2);
    memset(isBridge + 1, false, index);
    dfsClock = 0;
    tarjan(1, -1);
//shrink BCC to one node
    memset(low + 1, 0, N << 2);
    components = 0;
    for(int i = 1; i <= N; ++i){
        if(!low[i]){
            ++components;
            shrink(i);
        }
    }
//    printf("graph has %d components\n", components);
    if(components == 1){
        puts("0");
        return;
    }
//rebuild
    index = 0;
    memset(first + 1, -1, components << 2);
    for(int i = 0; i < M; ++i){
        int x = low[uu[i]], y = low[vv[i]];
        if(x == y) continue;
        addEdge(x, y);
        addEdge(y, x);
    }
//find all leaves
    int leaves = 0;
    for(int i = 1; i <= components; ++i){
        if(edge[first[i]].next == -1) ++leaves;
    }
    printf("%d\n", (leaves + 1) >> 1);
}
bool inputGraph()
{
    if(2 != scanf("%d%d", &N, &M)) return false;
    
    index = 0;
    memset(first + 1, -1, N << 2);
    
    for(int i = 0; i < M; ++i){
        scanf("%d%d", &uu[i], &vv[i]);
        addEdge(uu[i], vv[i]);
        addEdge(vv[i], uu[i]);
    }
    return true;
}

int main()
{
    while(inputGraph()) solve();
    return 0;
}