标签：
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.
------------------------------------------------------------------
#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
[Submit][Status][Discuss]Description
Input
Output
Sample Input
7 6
1 2
2 3
5 4
4 3
3 6
6 7
Sample Output
2
HINT
Source
BZOJ 2718: [Violet 4]毕业旅行( 最长反链 )
标签：
原文地址：http://www.cnblogs.com/JSZX11556/p/4853457.html
.gif)
.gif)
.gif)
