题目链接

题意：给出一张有向图，判断图上的任意两个点是否存在一条路可达(单向可达即可)。

思路：有向图找出强连通分量，然后缩点，因为题目要求任意两点存在可达的路，所以缩点之后的点，要形成一条单链，才能符合可达的要求，在这里用DP求最长路来判断是否能形成一条单链。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN = 2010;
const int MAXM = 50010;

struct Edge{
    int to, next;
}edge[MAXM];

int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];
int Index, top;
int scc;
bool Instack[MAXN];
int num[MAXN];
int n, m;
int d[MAXN];
vector<int> g[MAXN];

void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v) {
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void Tarjan(int u) {
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].to;         
        if (!DFN[v]) {
            Tarjan(v); 
            if (Low[u] > Low[v]) Low[u] = Low[v];
        } 
        else if (Instack[v] && Low[u] > DFN[v]) 
            Low[u] = DFN[v];
    }
    if (Low[u] == DFN[u]) {
        scc++; 
        do { 
            v = Stack[--top]; 
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        } while (v != u); 
    }
}

void solve() {
    memset(Low, 0, sizeof(Low));
    memset(DFN, 0, sizeof(DFN));
    memset(num, 0, sizeof(num));
    memset(Stack, 0, sizeof(Stack));
    memset(Instack, false, sizeof(Instack));
    Index = scc = top = 0;
    for (int i = 1; i <= n; i++) 
        if (!DFN[i])
            Tarjan(i);
}

int dp(int i) {
    int& ans = d[i];
    if (ans > 0) return ans;
    ans = 1;
    for (int j = 0; j < g[i].size(); j++) {
        int v = g[i][j]; 
        ans = max(ans, dp(v) + 1);
    }
    return ans;
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        scanf("%d%d", &n, &m); 
        int u, v;
        for (int i = 0; i < m; i++) {
            scanf("%d%d", &u, &v); 
            addedge(u, v); 
        } 
        solve();
    
        if (scc == 1) {
            printf("Yes\n"); 
            continue;
        }
        else {
            for (int i = 0; i <= scc; i++) g[i].clear(); 
            for (int u = 1; u <= n; u++) {
                for (int i = head[u]; i != -1; i = edge[i].next) {
                    int v = edge[i].to; 
                    if (Belong[u] != Belong[v]) 
                        g[Belong[u]].push_back(Belong[v]);
                } 
            } 
            memset(d, -1, sizeof(d));
            int ans = 0;
            for (int i = 1; i <= scc; i++)
                ans = max(ans, dp(i));
            if (ans == scc) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}

POJ2762-Going from u to v or from v to u?(强连通缩点+DP)