bnu 34982 Beautiful Garden(暴力)

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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;}

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