题意介绍
大学班级选班长,N 个同学均可以发表意见 若意见为 A B 则表示 A 认为 B 合适,意见具有传递性,即 A 认为 B 合适,B 认为 C 合适,则 A 也认为 C 合适,给出M个意见,找到得票最多的人
题意分析
dfs求出每个点所在的连通分量,然后缩点,并记录缩点之后每个点的入度(要将图反向),遍历得到每个点连通的点数,减去自己即为得票数,而与该点在同一个连通分量内的点都可选。
通过代码
#include<iostream>
#include<vector>
#include<algorithm>
#include<stack>
#include<string.h>
using namespace std;
int t, n, m;
const int MAXN = 5010;
const int inf = 0x3f3f3f3f;
int vis[MAXN], cmp[MAXN],pre[MAXN],low[MAXN],sz[MAXN],dfs_clock,k;
vector<int> G[MAXN], rG[MAXN];
stack<int> s;
void dfs(int u) {
pre[u] = low[u] = ++dfs_clock;
s.push(u);
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (!pre[v]) {
dfs(v);
low[u] = min(low[u], low[v]);
}
else if (!cmp[v]) low[u] = min(low[u], low[v]);
}
if (pre[u] == low[u]){
k++;
for (;;) {
int x = s.top();
s.pop();
cmp[x] = k;//记录每个点所在的连通分量编号
sz[k]++;
if (x == u) break;
}
}
}
int rdfs(int u) {
int ans = sz[u];
vis[u] = true;
for (int i = 0; i < rG[u].size(); i++) {
int v = rG[u][i];
if (vis[v]) continue;
ans += rdfs(v);
}
return ans;
}
void scc() {
dfs_clock = k = 0;
memset(pre, 0, sizeof(pre));
memset(cmp, 0, sizeof(cmp));
memset(sz, 0, sizeof(sz));
for (int i = 0; i < n; i++) {
if (pre[i] == 0)
dfs(i);
}
}
int in_deg[MAXN];
int support[MAXN];
int main() {
cin.sync_with_stdio(false);
cin >> t;
for(int ca=1;ca<=t;ca++){
cin >> n >> m;
for (int i = 0; i <= n; i++) {
G[i].clear();
rG[i].clear();
}
int x, y;
for (int i = 0; i < m; i++) {
cin >> x >> y;
G[x].push_back(y);
}
scc();
memset(in_deg, 0, sizeof(in_deg));
//缩点
for (int i = 0; i < n; i++) {
for (int j = 0; j < G[i].size(); j++) {
int u = cmp[i], v = cmp[G[i][j]];
if (u == v) continue;
in_deg[u]++;
rG[v].push_back(u);
}
}
int mx = 0;
for (int i = 1; i <= k; i++) {
if (in_deg[i] == 0) {
memset(vis, 0, sizeof(vis));
support[i] = rdfs(i);//记录每个点连通的点数
mx = max(mx, support[i]);//记录最大值
}
}
cout << "Case " << ca << ": " << mx-1 << endl;
bool flag = false;
for (int i = 0; i < n; i++) {
int u = cmp[i];
if (in_deg[u] == 0 && support[u] == mx) {
if (flag) cout << " ";
cout << i;
flag = true;
}
}
cout << endl;
}
return 0;
}