BZOJ4897 [Thu Summer Camp2016]成绩单 【dp】

题目链接

BZOJ4897

题解

发现我们付出的代价与区间长度无关，而与区间权值范围有关

离散化一下权值

我们设\(f[l][r][x][y]\)表示区间\([l,r]\)消到只剩权值在\([x,y]\)所需最小代价

\(f[l][r][0][0]\)即为消完的最小代价

那么

\[f[l][r][0][0] = min\{f[l][r][x][y] + a + b(w[y] - w[x])^2\}
\]

转移的话，贪心地取出区间两边在权值区间\([x,y]\)以内的数，剩下区间\([l',r']\)

如果剩余区间直接消去，可以直接计算

如果不一次消去，那么枚举断点转移即可

复杂度小常数\(O(n^5)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 55,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int f[maxn][maxn][maxn][maxn],n,a,b,w[maxn],c[maxn],tot;
void cmin(int& x,int y){x = min(x,y);}
int main(){
	n = read(); a = read(); b = read();
	REP(i,n) c[i] = w[i] = read();
	sort(c + 1,c + 1 + n); tot = 1;
	for (int i = 2; i <= n; i++) if (c[i] != c[tot]) c[++tot] = c[i];
	for (int i = 1; i <= n; i++) w[i] = lower_bound(c + 1,c + 1 + tot,w[i]) - c;
	for (int l = 1; l <= n; l++)
		for (int r = l; r <= n; r++){
			int mx = -INF,mn = INF;
			for (int k = l; k <= r; k++)
				mx = max(mx,w[k]),mn = min(mn,w[k]);
			f[l][r][0][0] = a + b * (c[mx] - c[mn]) * (c[mx] - c[mn]);
			for (int x = 1; x <= tot; x++)
				for (int y = x; y <= tot; y++)
					f[l][r][x][y] = INF;
		}
	for (int len = 1; len <= n; len++){
		for (int l = 1; l + len - 1 <= n; l++){
			int r = l + len - 1;
			for (int x = 1; x <= tot; x++)
				for (int y = x; y <= tot; y++){
					int ll = l,rr = r;
					while (ll <= r && w[ll] >= x && w[ll] <= y) ll++;
					while (rr >= ll && w[rr] >= x && w[rr] <= y) rr--;

					if (ll > rr) f[l][r][x][y] = 0;
					else if (ll == rr) f[l][r][x][y] = a;
					else {
						for (int k = ll; k < rr; k++){
							cmin(f[l][r][x][y],f[ll][k][x][y] + f[k + 1][rr][x][y]);
							cmin(f[l][r][x][y],f[ll][k][0][0] + f[k + 1][rr][x][y]);
							cmin(f[l][r][x][y],f[ll][k][x][y] + f[k + 1][rr][0][0]);
							cmin(f[l][r][x][y],f[ll][rr][0][0]);
						}
					}
				}
			for (int x = 1; x <= tot; x++)
				for (int y = x; y <= tot; y++)
					cmin(f[l][r][0][0],a + b * (c[y] - c[x]) * (c[y] - c[x]) + f[l][r][x][y]);
		}
	}
	printf("%d\n",f[1][n][0][0]);
	return 0;
}