Codeforces Round #371 (Div. 1) C - Sonya and Problem Wihtout a Legend
C - Sonya and Problem Wihtout a Legend
思路:感觉没有做过这种套路题完全不会啊。。 把严格单调递增转换成非严格单调递增,所有可能出现的数字就变成了原数组出现过的数字。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define PLI pair<LL, int> #define ull unsigned long long using namespace std; const int N = 3000 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; LL dp[N][N], a[N], hs[N], tot, n; int main() { memset(dp, INF, sizeof(dp)); scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%lld", &a[i]); a[i] = a[i] - i; hs[tot++] = a[i]; } sort(hs, hs + tot); tot = unique(hs, hs + tot) - hs; for(int j = 0; j < tot; j++) dp[1][j] = abs(a[1] - hs[j]); for(int i = 2; i <= n; i++) { LL mn = INF; for(int j = 0; j < tot; j++) { mn = min(mn, dp[i - 1][j]); dp[i][j] = min(dp[i][j], mn + abs(a[i] - hs[j])); } } LL ans = INF; for(int j = 0; j < tot; j++) ans = min(ans, dp[n][j]); printf("%lld\n", ans); return 0; } /* */