POJ - 2762 Going from u to v or from v to u?(拓扑排序+强连通分量)

2015年8月15日 17点热度 0条评论 来源: 暗金色

题目大意:给出N个点,M条有向边,问是否任意两点u,v都满足u能到达v或者v能到达u

解题思路:强连通分量内的所有的点都满足,接着要判断一下其他的点能否满足了
求出所有的强连通分量,接着缩点,用桥连接,形成新的图(以下所说的点都是指新的图的点)
如果一个点同时指向另外两个不同的点,那么这两个点之间肯定是不能相互到达的,所以拓扑排序一下,就可以知道是否符合了

#include <cstdio>
#include <cstring>

#define min(a,b) ((a) < (b) ? (a) : (b))
#define N 10100
#define M 60100

struct Edge{
    int from, to, next;
}E[M];

struct Node{
    int x, y;
}node[M];

int n, m, tot, dfs_clock, top, scc_cnt;
int head[N], pre[N], stack[N], sccno[N], lowlink[N], in[N];

void AddEdge(int u, int v) {
    E[tot].from = u;
    E[tot].to = v;
    E[tot].next = head[u];
    head[u] = tot++;
}

void init() {
    scanf("%d%d", &n, &m);

    memset(head, -1, sizeof(head));
    tot = 0;

    int u, v;
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &node[i].x, &node[i].y);
        AddEdge(node[i].x, node[i].y);
    }
}

void dfs(int u) {
    pre[u] = lowlink[u] = ++dfs_clock;
    stack[++top] = u;

    int v;
    for (int i = head[u]; i != -1; i = E[i].next) {
        v = E[i].to;
        if (!pre[v]) {
            dfs(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if (!sccno[v]) {
            lowlink[u] = min(lowlink[u], pre[v]);
        }
    }

    if (pre[u] == lowlink[u]) {
        scc_cnt++;
        while (1) {
            v = stack[top--];
            sccno[v] = scc_cnt;
            if (v == u) break;
        }
    }
}

int find() {
    int cnt = 0, num;
    for (int i = 1; i <= scc_cnt; i++) {
        if (in[i] == 0) {
            cnt++;
            num = i;
        }
    }
    if (cnt == 1)
        return num;
    return 0;
}

bool TopoOrder(){
    int cnt = 0;
    int u;
    while (u = find()) {
        in[u] = -1;
        for (int i = head[u]; i != -1; i = E[i].next) {
            in[E[i].to]--;
        }
        cnt++;
    }
    return cnt == scc_cnt;
}

void solve() {
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    dfs_clock = top = scc_cnt = 0;

    for (int i = 1; i <= n; i++)
        if (!pre[i])
            dfs(i);

    if (scc_cnt == 1) {
        printf("Yes\n");
        return ;
    }

    memset(in, 0, sizeof(in));
    memset(head, -1, sizeof(head));
    tot = 0;

    int u, v;
    for (int i = 0; i < m; i++) {
        u = sccno[node[i].x];
        v = sccno[node[i].y];
        if (u != v) {
            in[v]++;
            AddEdge(u, v);
        }
    }
    int ans = 0;
    for (int i = 1; i <= scc_cnt; i++)
        if (in[i] == 0)
            ans++;
    if (ans > 1) {
        printf("No\n");
        return ;
    }

    if (TopoOrder()) 
        printf("Yes\n");
    else
        printf("No\n");
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}
    原文作者:暗金色
    原文地址: https://blog.csdn.net/L123012013048/article/details/47673605
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