csu 2168: Fixed Point

题解：先把m次操作的置换给求出来，记住每个点在该置换中的位置，长度不同的置换最多有sqrt（n）个数，这里我自己只开了300个（够用）。然后对每个询问按%m分类，然后对每个%m类 求出每个点贡献，再统计答案。

#include"bits/stdc++.h"
using namespace std;
const int MX = 1e5+7;
int n,m,q;
int res[MX][300];
int ex[MX],ex2[MX];
int l[11],r[11];
int ans[MX];
int len[MX],blo[MX],pos[MX],col,maxl;
vector<int> G[MX];
struct node{
    int k,id;
    node(){}
    node(int k, int id) : k(k), id(id){}
};

vector<node> rem[11];
vector<int> Len;
map<int,int> mp;
bool vis[MX];
bool check(int *a, int *b)
{
    for(int i = 1; i <= n; i++)
        if(a[i] != b[i]) return 0;
    return 1;
}

void work(int mm)
{
    for(int i = 0; i <= maxl; i++)
        for(int j = 0; j < Len.size(); j++)
            res[i][j] = 0;

    for(int i = 1; i <= col; i++) {
        int sz = G[i].size();
        for(int j = 0; j < sz; j++) {
            int x = G[i][j], y = ex2[x], le = len[blo[x]];
            if(blo[x] != blo[y]) continue;
            int p = pos[x] - pos[y];
            if(p < 0) p += le;
            res[p][mp[le]]++;
        }
    }
    for(int i = 0; i < rem[mm].size(); i++) {
        int k = rem[mm][i].k, id = rem[mm][i].id;
        int ret = 0;
        for(int j = 0; j < Len.size(); j++) {
            int lx = Len[j];
            ret += res[(k/m)%lx][j];
        }
        ans[id] = ret;
    }
}

int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
    while(~scanf("%d%d%d",&n,&m,&q)) {
        assert(n >= 1 && n <= 100000 && m >= 1 && m <= 10 && q >= 1 && q <= 100000);
        mp.clear(); Len.clear();
        for(int i = 1; i <= n; i++) {
            ex2[i] = ex[i] = i;
            G[i].clear();
            vis[i] = 0;
        }
        for(int j = 1; j <= m; j++) {
            scanf("%d%d",&l[j],&r[j]);
        }

        for(int i = 1; i <= m; i++) {
            for(int j = l[i], l = r[i]; j < l; j++, l--) {
                swap(ex[j],ex[l]);
            }
        }

        int tyl = 0;
        maxl = col = 0;
        for(int i = 1; i <= n; i++) {
            int x = i, l = 0;
            if(vis[x]) continue;
            ++col;
            while(!vis[x]) {
                ++l;
                vis[x] = 1;
                blo[x] = col;
                pos[x] = l - 1;
                G[col].push_back(x);
                x = ex[x];
            }

            if(!mp.count(l)) {
                mp[l] = tyl++;
                Len.push_back(l);
            }
            len[blo[x]] = l;
            maxl = max(maxl,l);
        }

        for(int i = 1,k; i <= q; i++) {
            scanf("%d",&k);
            int re = k%m;
            rem[re].push_back(node(k,i));
        }
        for(int i = 0; i < m; i++) {
            work(i);
            rem[i].clear();
            for(int j = l[i+1], l = r[i+1]; j < l; j++,l--){
                swap(ex2[j],ex2[l]);
            }
        }

        for(int i = 1; i <= q; i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}