HDU 5009 Paint Pearls

题目链接

题意：给定一个目标颜色，每次能选一个区间染色，染色的代价为这个区间不同颜色数的平方，问最小代价

思路：先预处理，把相同颜色的一段合并成一个点，然后把颜色离散化掉，然后进行dp，dp[i]表示染到第i个位置的代价，然后往后转移，转移的过程记录下不同个数，这样就可以转移了，注意加个剪枝，就是如果答案大于了dp[n]就不用往后继续转移了

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 50005;

int n, vis[N];
vector<int> save;

struct Point {
	int val, id, rank;
} p[N];

bool cmpv(Point a, Point b) {
	return a.val < b.val;
}

bool cmpid(Point a, Point b) {
	return a.id < b.id;
}

const int INF = 0x3f3f3f3f;

int dp[N];

int main() {
	while (~scanf("%d", &n)) {
		for (int i = 1; i <= n; i++)
			scanf("%d", &p[i].val);
		int pn = 1;
		for (int i = 2; i <= n; i++) {
			if (p[i].val != p[i - 1].val)
				p[++pn] = p[i];
		}
		n = pn;
		for (int i = 1; i <= n; i++)
			p[i].id = i;
		sort(p + 1, p + n + 1, cmpv);
		p[1].rank = 0;
		int s = 0;
		for (int i = 2; i <= n; i++) {
			if (p[i].val != p[i - 1].val)
				++s;
			p[i].rank = s;
		}
		sort(p + 1, p + n + 1, cmpid);
		memset(dp, INF, sizeof(dp));
		dp[0] = 0;
		dp[n] = n;
		for (int i = 0; i < n; i++) {
			int cnt = 0;
			if (dp[i] > dp[i + 1]) continue;
			for (int j = i + 1; j <= n; j++) {
				if (vis[p[j].rank] == 0) {
					save.push_back(p[j].rank);
					cnt++;
				}
				vis[p[j].rank]++;
				if (dp[i] + cnt * cnt >= dp[n]) break;
				dp[j] = min(dp[j], dp[i] + cnt * cnt);
			}
			for (int i = 0; i < save.size(); i++)
				vis[save[i]] = 0;
			save.clear();
		}
		printf("%d\n", dp[n]);
	}
	return 0;
}