BZOJ1098 [POI2007]办公楼biu

终于考完了期中考。。。我活着回家了！

开始写题ing》》》》》

这道题嘛。。。先转化成补图，然后问题就变成求连通块个数。

但是会T，是因为点多而且是稠密图。。。

于是就用链表优化就好啦~

/**************************************************************
    Problem: 1098
    User: rausen
    Language: C++
    Result: Accepted
    Time:5708 ms
    Memory:34108 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
const int N = 100005;
const int M = 4000005;
 
struct edges{
    int next, to;
    edges() {}
    edges(int a, int b) : next(a), to(b) {}
}e[M];
 
int n, m, ans;
int first[N], tot;
int q[N], l, r;
int pre[N], next[N], a[N];
bool to[N];
 
inline int read(){
    int x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9')
        ch = getchar();
    while (ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
void add_Edges(int x, int y){
    e[++tot] = edges(first[x], y), first[x] = tot;
    e[++tot] = edges(first[y], x), first[y] = tot;
}
 
inline void del(int x){
    int tmp = pre[x];
    next[tmp] = next[x];
    pre[next[x]] = tmp;
}
 
void bfs(int p){
    q[0] = p;
    int now, x;
    for (l = r = 0; l <= r; ++l){
        ++a[ans];
        now = q[l];
        for (x = first[now]; x; x = e[x].next)
            to[e[x].to] = 1;
        for (x = next[0]; x <= n; x = next[x])
            if (!to[x])
                del(x), q[++r] = x;
        for (x = first[now]; x; x = e[x].next)
            to[e[x].to] = 0;
    }
}
 
int main(){
    n = read(), m = read();
    int i, x, y;
    for (i = 0; i <= n; ++i)
        next[i] = i + 1, pre[i + 1] = i;
    for (i = 1; i <= m; ++i){
        x = read(), y = read();
        add_Edges(x, y);
    }
    for (i = next[0]; i <= n; i = next[0])
        del(i), ++ans, bfs(i);
    printf("%d
", ans);
    sort(a + 1, a + ans + 1);
    for (i = 1; i <= ans; ++i)
        printf("%d ", a[i]);
    return 0;
}