题目链接：bnu 34982 Beautiful Garden

题目大意：给定一个长度为n的序列，问说最少移动多少点，使得序列成等差序列，点的位置可以为小数。

解题思路：算是纯暴力吧，枚举等差的起始和中间一点，因为要求定中间一点的位置，所以这一步是o(n3);然后用o(n)的算法确定说需要移动几个来保证序列等差。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
const int N = 1e5+5;

bool flag;
int n, m, c, mv, f[N], r[N], ans[N];
vector<int> g[N];

int getfar(int x) {
    return x == f[x] ? x : f[x] = getfar(f[x]);
}

void init () {
    scanf("%d%d", &n, &m);

    flag = false;
    for (int i = 0; i <= n; i++)
        r[i] = f[i] = i;
    for (int i = 0; i < m; i++)
        g[i].clear();

    int t;
    for (int i = 0; i < m; i++) {
        scanf("%d", &t);
        int a, pre = 0;

        for (int j = 0; j < t; j++) {
            scanf("%d", &a);
            g[i].push_back(a);
            if (a < pre)
                flag = true;
            pre = a;
        }
    }
}

bool insert (int x, int d) {

    for (int j = mv-1; j >= 0; j--) {
        if (g[d][j] < x) {
            int p = getfar(g[d][j]);
            f[p] = x;
            r[p] = x;
            mv = j;
            return true;
        }
    }
    return false;
}

void put(int x) {
    ans[c--] = x;
    if (r[x] != x)
        put(r[x]);
}

void solve () {

    for (int i = m-1; i; i--) {
        int t = g[i].size();

        mv = g[i-1].size();

        for (int j = t-1; j >= 0; j--)
            if (!insert(g[i][j], i-1)) {
                flag = true;
                return;
            }
    }

    c = n;
    int t = g[0].size();
    for (int i = t-1; i >= 0; i--)
        put(g[0][i]);
}

int main () {
    int cas; 
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        init ();

        printf("Case #%d: ", i);
        solve();

        if (flag) {
            printf("No solution\n");
        } else {
            for (int j = 1; j < n; j++)
                printf("%d ", ans[j]);
            printf("%d\n", ans[n]);
        }
    }
    return 0;
}