20240916总结

不积跬步，无以千里。

这两天主要是复习了图的连通性相关的题+听了youwike哥哥讲课。

先是复习了缩点，割点，割边，点双，边双，2-SAT，感觉比较需要注意的是割点的那个第一个节点的判断，写题的时候总是容易忘。然后又写了几道练习题。

• 缩点

#include <iostream>
#include <cstring>

using namespace std;
const int N = 1e5 + 10;

int n, m, tot, cnt, top, scc, ans;
int a[N], s[N], sta[N], x[N], y[N], f[N];
int head[N], dfn[N], low[N], col[N];
bool vis[N];
struct Map { int to, nxt; } e[N << 1];

void add(int u, int v) {
	e[++tot] = {v, head[u]};
	head[u] = tot;
}

void tarjan(int u) {
	low[u] = dfn[u] = ++cnt;
	vis[u] = 1, sta[++top] = u;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!dfn[v]) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if(vis[v]) 
			low[u] = min(low[u], dfn[v]);
	}
	if(dfn[u] == low[u]) {
        scc++;
        while(sta[top + 1] != u) {
            col[sta[top]] = scc;
            s[scc] += a[sta[top]];
            vis[sta[top--]] = 0;
        }
    }
}

void dfs(int x) {
	if(f[x]) return;
	f[x] = s[x];
	int maxx = 0;
	for(int i = head[x]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!f[v]) dfs(v);
		maxx = max(maxx, f[v]);
	}
	f[x] += maxx;
}

int main() {
	cin >> n >> m;
	for(int i = 1; i <= n; ++i) cin >> a[i];
	for(int i = 1, u, v; i <= m; ++i)
		cin >> x[i] >> y[i], add(x[i], y[i]);
	for(int i = 1; i <= n; ++i)
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= tot; ++i) e[i] = {0, 0};
	for(int i = 1; i <= n; ++i) head[i] = 0;
	tot = 0;
	for(int i = 1; i <= m; ++i) 
		if(col[x[i]] != col[y[i]]) add(col[x[i]], col[y[i]]);
	for(int i = 1; i <= scc; ++i) {
		if(!f[i]) 
			dfs(i), ans = max(ans, f[i]);
	}
	cout << ans << endl;
	return 0;
}

• 点双

#include <iostream>
#include <vector>

using namespace std;
const int N = 4e6 + 10;

int n, m, tot, top, cnt, scc;
int s[N];
int head[N], dfn[N], low[N];
vector <int> ans[N];
struct Map { int to, nxt; } e[N << 1];

void add(int u, int v) {
	e[++tot] = {v, head[u]};
	head[u] = tot;
}

void tarjan(int u, int fa) {
	int son = 0;
	low[u] = dfn[u] = ++cnt;
	s[++top] = u;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!dfn[v]) {
			son++, tarjan(v, u);
			low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]) {
				scc++;
				while(s[top + 1] != v) ans[scc].push_back(s[top--]);
				ans[scc].push_back(u);
			}
		} else if(v != fa) low[u] = min(low[u], dfn[v]);
	}
	if(fa == 0 && son == 0) ans[++scc].push_back(u);
}

int main() {
	cin >> n >> m;
	for(int i = 1, u, v; i <= m; ++i)
		cin >> u >> v, add(u, v), add(v, u);
	for(int i = 1; i <= n; ++i)
		if(!dfn[i]) top = 0, tarjan(i, 0);
	cout << scc << endl;
	for(int i = 1; i <= scc; ++i) {
		cout << ans[i].size() << ' ';
		for(int j = 0; j < ans[i].size(); ++j) 
			cout << ans[i][j] << ' ';
		cout << endl;
	}
	return 0;
}

• 2-SAT

#include <iostream>

using namespace std;
const int N = 2e6 + 10;

int n, m, tot, cnt, scc, top;
int head[N], dfn[N], low[N];
int s[N], col[N];
bool vis[N];
struct Map { int to, nxt; } e[N << 1];

void add(int u, int v) {
	e[++tot] = {v, head[u]};
	head[u] = tot;
}

void tarjan(int u) {
	low[u] = dfn[u] = ++cnt;
	vis[u] = 1, s[++top] = u;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!dfn[v]) 
			tarjan(v), low[u] = min(low[u], low[v]);
		else if(vis[v]) low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u]) {
		++scc;
		while(u != s[top + 1]) {
			col[s[top]] = scc;
			vis[s[top--]] = 0;
		}
	}
} 

int main() {
	cin >> n >> m;
	for(int i = 1, u, v, vu, vv; i <= m; ++i) {
		cin >> u >> vu >> v >> vv;
		add(u + n * (vu & 1), v + n * (vv ^ 1));
		add(v + n * (vv & 1), u + n * (vu ^ 1));
	}
	for(int i = 1; i <= (n << 1); ++i) 
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= n; ++i)
		if(col[i] == col[n + i]) 
			return cout << "IMPOSSIBLE" << endl, 0;
	cout << "POSSIBLE" << endl;
	for(int i = 1; i <= n; ++i)
		cout << (col[i] < col[n + i]) << ' ';
	return 0;
}

• 满汉全席
  把$m$和$h$看成两个对立的限制然后直接像2-SAT一样建图就行了，要注意的是对于每组数据都要全部初始化。

#include <iostream>
#include <cstring>

using namespace std;
const int N = 2e6 + 10;

int T, n, m, tot, cnt, scc, top;
int head[N], dfn[N], low[N];
int s[N], col[N];
char s1[110], s2[110];
bool vis[N];
struct Map { int to, nxt; } e[N << 1];

void add(int u, int v) {
	e[++tot] = {v, head[u]};
	head[u] = tot;
}

void tarjan(int u) {
	low[u] = dfn[u] = ++cnt;
	vis[u] = 1, s[++top] = u;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!dfn[v]) 
			tarjan(v), low[u] = min(low[u], low[v]);
		else if(vis[v]) low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u]) {
		++scc;
		while(u != s[top + 1]) {
			col[s[top]] = scc;
			vis[s[top--]] = 0;
		}
	}
} 

void Main() {
	cin >> n >> m;
	for(int i = 1; i <= tot; ++i) e[i] = {0, 0};
	memset(head, 0, sizeof(head)), memset(dfn, 0, sizeof(dfn));
	memset(low, 0, sizeof(low)), memset(s, 0, sizeof(s));
	memset(col, 0, sizeof(col)), memset(vis, 0, sizeof(vis));
	tot = cnt = scc = top = 0;
	for(int i = 1, u, v, vu, vv; i <= m; ++i) {
		cin >> s1 >> s2, u = v = 0;
		int lu = strlen(s1), lv = strlen(s2);
		vu = (s1[0] == 'm'), vv = (s2[0] == 'm');
		for(int j = 1; j < lu; ++j) u = u * 10 + s1[j] - '0';
		for(int j = 1; j < lv; ++j) v = v * 10 + s2[j] - '0';
		add(u + n * (vu & 1), v + n * (vv ^ 1));
		add(v + n * (vv & 1), u + n * (vu ^ 1));
	}
	for(int i = 1; i <= (n << 1); ++i) 
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= n; ++i)
		if(col[i] == col[n + i]) 
			return cout << "BAD" << endl, void();
	cout << "GOOD" << endl;
}

int main() {
	cin >> T;
	while(T--) Main();
	return 0;
}

• Redundant Paths G
  因为是双向边，所以缩点之后就肯定是一棵树，然后要至少度数是2，所以最好的方法就是每两个缩点之后的叶子节点连边，所以答案就是$(s-1)/2$(s是叶子节点个数)

#include <iostream>
#include <vector>

using namespace std;
const int N = 2e6 + 10;

int n, m, tot = 1, top;
int cnt, scc, ans;
int head[N], u[N], v[N], d[N];
int dfn[N], low[N], s[N], col[N];
bool vis[N];
struct Map { int to, nxt; bool flag; } e[N * 2];

void add(int u, int v) {
	e[++tot] = {v, head[u], 0};
	head[u] = tot;
}

void tarjan(int u) {
	low[u] = dfn[u] = ++cnt;
	s[++top] = u;
	for(int i = head[u]; i; i = e[i].nxt) {
		if(vis[i]) continue;
		vis[i] = vis[i ^ 1] = 1;
		int v = e[i].to;
		if(!dfn[v]) 
			tarjan(v), low[u] = min(low[u], low[v]);
		else low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u]) {
		++scc;
		while(u != s[top + 1]) col[s[top--]] = scc;
	}
}

int main() {
	cin >> n >> m;
	for(int i = 1; i <= m; ++i) 
		cin >> u[i] >> v[i], add(u[i], v[i]), add(v[i], u[i]);
	for(int i = 1; i <= n; ++i)
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= m; ++i) 
		if(col[u[i]] != col[v[i]]) d[col[u[i]]]++, d[col[v[i]]]++;
	for(int i = 1; i <= scc; ++i) 
		if(d[i] == 1) ans++;
	cout << (ans + 1) / 2 << endl;
	return 0;
}

• Running In The Sky
  缩点板子题，就多维护一个路径最大值就行了。

#include <iostream>
#include <vector>

using namespace std;
const int N = 2e6 + 10;

int n, m, tot = 1, top;
int cnt, scc, ans;
int head[N], u[N], v[N], d[N];
int dfn[N], low[N], s[N], col[N];
bool vis[N];
struct Map { int to, nxt; bool flag; } e[N * 2];

void add(int u, int v) {
	e[++tot] = {v, head[u], 0};
	head[u] = tot;
}

void tarjan(int u) {
	low[u] = dfn[u] = ++cnt;
	s[++top] = u;
	for(int i = head[u]; i; i = e[i].nxt) {
		if(vis[i]) continue;
		vis[i] = vis[i ^ 1] = 1;
		int v = e[i].to;
		if(!dfn[v]) 
			tarjan(v), low[u] = min(low[u], low[v]);
		else low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u]) {
		++scc;
		while(u != s[top + 1]) col[s[top--]] = scc;
	}
}

int main() {
	cin >> n >> m;
	for(int i = 1; i <= m; ++i) 
		cin >> u[i] >> v[i], add(u[i], v[i]), add(v[i], u[i]);
	for(int i = 1; i <= n; ++i)
		if(!dfn[i]) tarjan(i);
	for(int i = 1; i <= m; ++i) 
		if(col[u[i]] != col[v[i]]) d[col[u[i]]]++, d[col[v[i]]]++;
	for(int i = 1; i <= scc; ++i) 
		if(d[i] == 1) ans++;
	cout << (ans + 1) / 2 << endl;
	return 0;
}

• 菜肴制作
  容易发现，如果直接按照限制做的话是不对的，因为不是要最后字典序最小，而是要当前没有输出的数尽量靠前。但是可以发现越大的书越靠后肯定是不劣的，所以只用倒着拓扑就行了。

#include <iostream>
#include <vector>
#include <queue>

using namespace std;
const int N = 1e5 + 10;

int T, n, m, cnt;
int in[N], ans[N];
vector <int> v[N];
priority_queue <int> q;

void Main() {
	cin >> n >> m;
	cnt = 0;
	for(int i = 1; i <= n; ++i)
		in[i] = ans[i] = 0, v[i].clear();
	for(int i = 1, u, vv; i <= m; ++i) {
		cin >> u >> vv;
		in[u]++, v[vv].push_back(u);
	}
	for(int i = 1; i <= n; ++i)
		if(!in[i]) q.push(i);
	while(!q.empty()) {
		int u = q.top();
		q.pop();
		for(int i = 0; i < v[u].size(); ++i) {
			int y = v[u][i];
			in[y]--;
			if(!in[y]) q.push(y);
		}
		ans[++cnt] = u;
	}
	if(cnt < n) return cout << "Impossible!" << endl, void();
	for(int i = cnt; i; --i) cout << ans[i] << ' ';
	cout << endl;
}

int main() {
	cin >> T;
	while(T--) Main();
	return 0;
}

感觉缩点很好使啊，就是考试没见过

__EOF__

本文作者：Rose_Lu
本文链接：https://www.cnblogs.com/roselu/p/18416600.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！