luogu P1330 封锁阳光大学

Posted 2021-01-16 mrclr

嘟嘟嘟

 

又刷了一道水题……

很显然只要判断这个图是否是二分图就行了，判断方法就是染色。如果对于边(u->v)，两个点颜色相同，那么就说明图中存在奇环，不是二分图。

统计答案的时候输出两种颜色较小的就行了。

需要注意的是，图可能不连通，或者有些点不存在。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("") 
13 #define space putchar(‘ ‘)
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 const int maxn = 1e4 + 5;
21 const int maxe = 1e5 + 5;
22 inline ll read()
23 {
24   ll ans = 0;
25   char ch = getchar(), last = ‘ ‘;
26   while(!isdigit(ch)) last = ch, ch = getchar();
27   while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
28   if(last == ‘-‘) ans = -ans;
29   return ans;
30 }
31 inline void write(ll x)
32 {
33   if(x < 0) x = -x, putchar(‘-‘);
34   if(x >= 10) write(x / 10);
35   putchar(x % 10 + ‘0‘);
36 }
37 
38 int n, m, ans = 0;
39 struct Edge
40 {
41   int nxt, to;
42 }e[maxe << 1];
43 int head[maxn], ecnt = -1;
44 void addEdge(int x, int y)
45 {
46   e[++ecnt] = (Edge){head[x], y};
47   head[x] = ecnt;
48 }
49 
50 int vis[maxn];
51 int sta[maxn], top = 0;
52 void dfs(int now, int col)
53 {
54   sta[++top] = now;
55   vis[now] = col; 
56   for(int i = head[now]; i != -1; i = e[i].nxt)
57     {
58       if(vis[e[i].to] == vis[now]) {puts("Impossible"); exit(0);}
59       if(vis[e[i].to]) continue;
60       dfs(e[i].to, col ^ 1);
61     }
62 }
63 
64 int main()
65 {
66   Mem(head, -1);
67   n = read(); m = read();
68   for(int i = 1; i <= m; ++i)
69     {
70       int x = read(), y = read();
71       addEdge(x, y); addEdge(y, x);
72     }
73   for(int i = 1; i <= n; ++i)
74     {
75       if(!vis[i])
76     {
77       dfs(i, 2);
78       int ans1 = 0, ans2 = 0;
79       while(top)
80         {
81           ans1 += (vis[sta[top]] == 2);
82           ans2 += (vis[sta[top]] == 3);
83           top--;
84         }
85       ans += min(ans1, ans2);
86     }
87     }
88   write(ans), enter;
89   return 0;
90 }

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