Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B
Cactus
Input: standard input
Output: standard output

A directed graph is a set V of vertices and a set of E ∈ {V x V} edges. An edge (u,v) is said to be directed from u to v (the edge (v,u) has the opposite direction). A directed cycle in a directed graph is a sequence of edges

(u₁, v₁), (u₂, v₂),…, (u_k, v_k)

such that u_i+1 = v_i for i = 1, …, k-1, and u₁=v_k. The directed cycle is simple if u_i ≠ u_j whenever i ≠ j (i.e., if it does not pass through a vertex twice).

In a strongly connected directed graph, there is for every pair u,v of vertices some directed cycle (not necessarily simple) that visits both u and v.

A directed graph is a cactus if and only if it is strongly connected and each edge is part of exactly one directed simple cycle. The first graph is a cactus, but the second one is not since for instance the edge (0,1) is in two simple cycles.

The reason for the name is that a "cactus" consists of several simple cycles connected to each other in a tree-like fashion, making it look somewhat like a cactus.

Problem

Write a program that given a directed graph determines if it is a cactus or not. The graph can have several thousand vertices.

Input

The first line contains an integer which is the number of test cases (less than 20). Each test case starts a line with an integer n ≥ 0 followed by line with an integer m ≥ 0 giving the number of vertices (n) and edges (m) in a graph (at most 10,000 of each). The vertices are numbered 0 through n-1. The following m lines describe the edges as pairs of numbers u v denoting an edge directed from u to v. There will never be more than one edge from u to v for any pair of vertices u and v. There are no loops, i.e., no edges from a vertex to itself.

Output

For each test case output a single line with a single string. Output "YES" if the graph is a cactus, and output "NO" if it is not.

Sample Input

2
4
5
0 1
1 2
2 0
2 3
3 2
4
5
0 1
1 2
2 3
3 0
1 3

Sample Output

YES
NO

---

Problem setter: Mikael Goldmann

題目描述：

判斷這張圖是否會符合，恰好是一個 SCC 元件，以及每一條邊都只出現過一次於一個環上。

題目解法：

一樣使用 SCC 的遞迴樹，對於 back edge 時，將整個環標記。

因此不同於一般 SCC 寫法，需要多紀錄 parent。

而有些邊可能會使用多次，由於只看 back edge，多餘向下也肯定是多餘的邊。

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <deque>
#include <algorithm>
using namespace std;
struct edge {
    int to, cnt;
    edge(int a=0, int b=0):
        to(a) , cnt(b) {}
};
vector<edge> g[10005];
int vfind[10005], findIdx;
int stk[10005], stkIdx;
int in_stk[10005], visited[10005];
//
int parent[10005];
edge *parentUsed[10005];
int checkflag;
int scc(int nd) {
    in_stk[nd] = visited[nd] = 1;
    stk[++stkIdx] = nd;
    vfind[nd] = ++findIdx;
    int mn = vfind[nd], i;
    for(i = 0; i < g[nd].size(); i++) {
        if(!visited[g[nd][i].to]) {
            parent[g[nd][i].to] = nd;//this problem process
            parentUsed[nd] = &g[nd][i];//this problem process
            mn = min(mn, scc(g[nd][i].to));
        }
        if(in_stk[g[nd][i].to]) {
            mn = min(mn, vfind[g[nd][i].to]);
            if(vfind[g[nd][i].to] <= vfind[nd]) {//this problem process
                // cycly find.
                if(checkflag)    continue;
                g[nd][i].cnt++;
                if(g[nd][i].cnt > 1)    checkflag = 1;
                int next = nd;
                while(checkflag == 0) {
                    parentUsed[parent[next]]->cnt++;
                    if(parentUsed[parent[next]]->cnt > 1)    checkflag = 1;
                    next = parent[next];
                    if(next == g[nd][i].to)
                        break;
                }
            }
        }
    }
    if(mn == vfind[nd]) {
        do {
            in_stk[stk[stkIdx]] = 0;
        } while(stk[stkIdx--] != nd);
    }
    return mn;
}
int main() {
    int testcase;
    int n, m;
    int i, j, k, x, y;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &n, &m);
        for(i = 0; i < n; i++)
            g[i].clear();
        for(i = 0; i < m; i++) {
            scanf("%d %d", &x, &y);
            g[x].push_back(edge(y, 0));
        }
        memset(visited, 0, sizeof(visited));
        memset(in_stk, 0, sizeof(in_stk));
        int component = 0;
        checkflag = 0;
        for(i = 0; i < n; i++) {
            if(visited[i] == 0) {
                findIdx = stkIdx = 0;
                if(++component > 1)    break;
                scc(i);
            }
        }
        for(i = 0; i < n; i++)
            for(j = 0; j < g[i].size(); j++)
                if(g[i][j].cnt != 1)
                    checkflag = 1;
        if(component > 1)    checkflag = 1;
        puts(checkflag == 0 ? "YES" : "NO");
#define debug 0
#if debug == 1
        for(i = 0; i < n; i++)
            for(j = 0; j < g[i].size(); j++)
                printf("%d -> %d %d\n", i, g[i][j].to, g[i][j].cnt);
        printf("%d %d\n", checkflag, component);
#endif
    }
    return 0;
}