题意:给你n个数字,修改这些数字,以达到整个序列为非严格单调上升或下降的序列。
求最少修改代价,代价为每个数字修改前后的差值。
如1 3 2 4 5 3 9, 让第2个数字3修改为2,代价为3-2=1,倒数第2个3修改为5,代价为5-3=2,总代价为2+1=3,所以最终的序列为 1 2 2 4 5 5 9,非严格上升序列
AC代码:
#include <cstdio> #include <iostream> #include <cmath> #include <cstring> #include <algorithm> using namespace std; const int N = 2005; const int INF = 0x3f3f3f3f; int d[N][N],n,a[N],b[N],mx; //d[i][j] 前i个数的序列,以第j个数为最后一个数的最小代价 void solve() { int ans = INF; sort(b+1,b+n+1); memset(d, 0, sizeof(d)); for(int i = 1; i <= n; i++) { int mn = d[i-1][1]; for(int j = 1; j <= n; j++) { mn = min(mn, d[i-1][j]); d[i][j] = mn + abs(b[j]-a[i]); } } for(int i = 1; i <= n; i++) ans = min(ans, d[n][i]); printf("%d ", ans); } int main() { while(~scanf("%d", &n)) { mx = 0; for(int i = 1; i <= n; i++) scanf("%d", a+i),b[i]=a[i]; solve(); } return 0; }
AC代码(滚动数组):
#include <cstdio> #include <iostream> #include <cmath> #include <cstring> #include <algorithm> using namespace std; const int N = 2005; const int INF = 0x3f3f3f3f; int d[2][N],n,a[N],b[N],mx; //d[i][j] 前i个数的序列,以第j个数为最后一个数的最小代价 void solve() { int ans = INF; sort(b+1,b+n+1); memset(d, 0, sizeof(d)); int f = 0; for(int i = 1; i <= n; i++) { int mn = d[f][1]; for(int j = 1; j <= n; j++) { mn = min(mn, d[f][j]); d[!f][j] = mn + abs(b[j]-a[i]); } f = !f; } for(int i = 1; i <= n; i++) ans = min(ans, d[f][i]); printf("%d ", ans); } int main() { while(~scanf("%d", &n)) { mx = 0; for(int i = 1; i <= n; i++) scanf("%d", a+i),b[i]=a[i]; solve(); } return 0; }