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最新推荐文章于 2022-10-07 09:37:55 发布

转载最新推荐文章于 2022-10-07 09:37:55 发布 · 643 阅读

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#强连通

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本文介绍了一种基于强连通分量(SCC)的算法来计算无向图中的桥的数量。通过使用深度优先搜索(DFS)并结合栈来追踪递归调用，该算法能够有效地找出所有桥并将结果用于进一步的分析。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;

const int M = 1005;
struct node {

    int to;
    int next;
}num[M*3];
stack<int>s;
int head[M];
int ins[M];
int dfn[M];
int low[M];
int sccf[M];
int du[M];
int n, m;
int index, scc;
int e ;

void init() {

    index = 1;
    scc = 0;
    memset(head, -1, sizeof(head));
    memset(ins, 0, sizeof(ins));
    memset(low, 0, sizeof(low));
    memset(sccf, 0, sizeof(sccf));
    memset(du, 0, sizeof(du));
    memset(dfn, 0, sizeof(dfn));

}

void add(int a, int b) {

    num[e].to = b;
    num[e].next = head[a];
    head[a] = e++;
}

int getqiaoshu() {

 // printf("d %d\n", scc);
    int  v;
    for(int u = 1; u <= n; u++) {

        for(int k = head[u]; k != -1; k = num[k].next) {

            v = num[k].to;
            if(low[u] != low[v]) {

                du[low[u]]++;
            }
        }
    }
    int cont = 0;
    for(int i = 1; i <= n; i++) {

        if(du[i] == 1) {

            cont++;
        }
    }

    return (cont + 1)/2;
}

void Tanjian(int u, int p) {

    int v;
    low[u] = dfn[u] = index++;
    ins[u] = 1;
    s.push(u);
   for(int k = head[u]; k != -1; k = num[k].next) {

        v = num[k].to;
        if(v == p)  // 防止 a ->b 和 b -> a
            continue;
        if(!dfn[v]) {

            Tanjian(v, u);
            low[u] = min(low[v], low[u]);
        }
        else if(ins[v] == 1) {

            low[u] = min(low[v], low[u]);
        }

   }
    // ins[v] = 2;
   /*if(low[u] == dfn[u]) {

        scc++;
        do {

            v = s.top();
            s.pop();
            sccf[v] = scc;
            ins[v] = 0;
        }while(v != u);

   }*/

}

int main()
{
    int q = 1;
    int a, b;
    while(scanf("%d%d", &n, &m) != EOF) {

      //  printf("Sample Input %d\n", q++);

        init();
          e = 0;
        for(int i = 1; i <= m; i++) {

            scanf("%d%d", &a, &b);
            add(a, b);
            add(b, a);
        }
        for(int i = 1; i <= n; i++) {

            if(!dfn[i]) {

                Tanjian(i, i);
            }
        }

        int ans = getqiaoshu();
        printf("%d\n", ans);
    }


    return 0;
}