codeforces 402B - Trees in a Row

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题目链接：http://codeforces.com/problemset/problem/402/B

题目大意：给出这n棵树的高度，通过增加或是减少树的高度使得第i棵树比第i-1棵树高k米，求最小的步数及每步的操作。

题目分析1：枚举以哪棵树为基准。

代码参考1：

#include<map>#include<set>#include<cmath>#include<queue>#include<stack>#include<cstdio>#include<string>#include<cstring>#include<sstream>#include<iostream>#include<algorithm>#include<functional>using namespace std;const int N = 2000;int a[N], b[N];int main(){int n, k, i, j;while (~scanf("%d%d", &n, &k)){memset(b, 0, sizeof(b));for (i = 0; i < n; ++i){scanf("%d", &a[i]);}for (i = 0; i < n; ++i)//枚举基准{for (j = n - 1; j >= 0; --j){int d = j - i;if (a[j] - a[i] != d * k)//如果他们的高度差不符合条件，改变次数+1{b[i]++;}}if (a[i] - i * k <= 0)//a[0]是最小的数，如果它<=0的话就不符合条件，赋值为INF{b[i] = N;}}int m = N, p;for (i = 0; i < n; ++i)//选出修改次数最小的数和所在位置{if (b[i] < m){m = b[i];//最小次数p = i;//所在位置}}printf("%d\n", m);for (j = 0; j < n; ++j)//输出每步的具体操作{int d = a[j] - a[p] - (j - p - 1) * k;if (d > k){printf("- %d %d\n", j + 1, d - k);}elseif (d < k){printf("+ %d %d\n", j + 1, k - d);}}}return 0;}

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代码参考2：（小土豆的~）

#include<map>#include<set>#include<cmath>#include<queue>#include<stack>#include<cstdio>#include<string>#include<cstring>#include<sstream>#include<iostream>#include<algorithm>#include<functional>using namespace std;int cnt[1 << 10];int arr[1 << 10];int main(){int n, k, i, a;while (~scanf("%d%d", &n, &k)){memset(cnt, 0, sizeof(cnt));for (i = 1; i <= n; ++i){scanf("%d", &a);arr[i] = a;if (a - (i - 1)*k > 0){++cnt[a - (i - 1)*k];//a-(i-1)*k以i为基准1位置该有的高度//5 1//1 1 1 2 3//能让首项为1的有2个，能让首项为2或3的都只有1个，所以首项就是1了}}int ans = 0, base = -1;for (i = 0; i < 1<<10; ++i)//找出能让首项为i的次数最多的{if (cnt[i] > ans){ans = cnt[i];base = i;}}printf("%d\n", n - ans);for (i = 1; i <= n; ++i){if (base != arr[i]){if (base < arr[i]){printf("-");}else{printf("+");}printf(" %d %d\n", i, abs(base - arr[i]));}base += k;}}return 0;}

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