BZOJ 2718: [Violet 4]毕业旅行( 最长反链 )

一不小心速度就成了#1....

这道题显然是求最长反链, 最长反链=最小链覆盖.最小链覆盖就是先做一次floyd传递闭包, 再求最小路径覆盖. 最小路径覆盖=N - 二分图最大匹配. 所以把所有点拆成x,y两个, 然后存在edge(u,v)就连ux->vy. 然后跑匈牙利即可.

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#include<bits/stdc++.h> using namespace std; const int maxn = 209; inline int read() { char c = getchar(); int ret = 0; for(; !isdigit(c); c = getchar()); for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0'; return ret;} struct edge { int to; edge* next;} E[maxn * maxn], *pt = E, *head[maxn]; inline void addedge(int u, int v) { pt->to = v; pt->next = head[u]; head[u] = pt++;} bitset<maxn> d[maxn];int vis[maxn], match[maxn], N, C; bool dfs(int x) { for(edge* e = head[x]; e; e = e->next) if(vis[e->to] != C) { vis[e->to] = C; if(!~match[e->to] || dfs(match[e->to])) { match[e->to] = x; return true; } } return false;} int main() { memset(match, -1, sizeof match); memset(vis, -1, sizeof vis); N = read(); int m = read(); for(int i = 0; i < N; i++) d[i].reset(); while(m--) d[read() - 1][read() - 1] = 1; for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) if(d[j][i]) d[j] |= d[i]; for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) if(d[i][j]) addedge(i, j); int ans = N; for(C = 0; C < N; C++) if(dfs(C)) ans--; printf("%d\n", ans); return 0;}------------------------------------------------------------------

2718: [Violet 4]毕业旅行

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 317 Solved: 178
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Description

Input

Output

最多可选多少景点

Sample Input

7 6
1 2
2 3
5 4
4 3
3 6
6 7

Sample Output

2

HINT

Source

Ctsc2008 River & ural 1533. Fat Hobbits