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 = + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ; LL dp[N][N], a[N], hs[N], tot, n; int main() {
memset(dp, INF, sizeof(dp));
scanf("%d", &n);
for(int i = ; 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 = ; j < tot; j++) dp[][j] = abs(a[] - hs[j]); for(int i = ; i <= n; i++) {
LL mn = INF;
for(int j = ; j < tot; j++) {
mn = min(mn, dp[i - ][j]);
dp[i][j] = min(dp[i][j], mn + abs(a[i] - hs[j]));
}
}
LL ans = INF;
for(int j = ; j < tot; j++)
ans = min(ans, dp[n][j]);
printf("%lld\n", ans);
return ;
} /*
*/