{题解}[jzoj3864]【JSOI2014】歌剧表演

传送门

Analysis

由题，先把所有数看作一个集合
于是，每当”演出”时 可以将”演出”中的数与其他数字”分离”开来
显然 对于”单个”的数字 我们可以认出
于是 问题就解决了
重在思考的角度

Code

#include 
#include 
#include 
#include 
using namespace std;
const int N = 100100;
int n,m;
int id[N],a[N],pos[N],col[N],le[N],ri[N],ans[N];
bool cmp(int x, int y)
{
    return col[x] < col[y];
}
int main()
{
    //freopen("3864.in","r",stdin);
    scanf("%d%d", &n, &m);
    for (int i = 1;i <= n;i ++)
        id[i] = i, pos[i] = i, col[i] = 1;

    le[1] = 1,ri[1] = n;
    int tot = 1;
    for (int o = 1;o <= m;o ++)
    {
        int t;
        scanf("%d", &t);
        for (int i = 1;i <= t;i ++) scanf("%d", &a[i]);

        sort(a + 1,a + 1 + t,cmp);
        for (int l  = 1;l <= t;/*l ++*/) 
        {
            int r = l;
            while (col[a[r + 1]] == col[a[l]] && r < t) r ++;
            int now = col[a[l]];
            tot ++;
            for (int i = le[now],j = l;j <= r;i ++,j ++)
            {
                swap(id[i],id[pos[a[j]]]);
                pos[id[pos[a[j]]]] = pos[a[j]],
                pos[a[j]] = i,
                col[a[j]] = tot;
            }
            le[tot] = le[now],
            ri[tot] = le[tot] + r - l,
            le[now] = ri[tot] + 1;
            if (le[tot] == ri[tot] && !ans[id[le[tot]]]) ans[id[le[tot]]] = o;
            if (le[now] == ri[now] && !ans[id[le[now]]]) ans[id[le[now]]] = o;
            l = r + 1;
        }
    }
    for (int i = 1;i <= n;i ++) printf("%d ", ans[i]);
}