题意：健♂身♂房有两个仰卧起坐坐垫，每次调整角度要花费10元/10度，每次使用要花费15，如今给定n个人的时间顺序，和所希望的角度，求最少花费

思路：dp，dp[i][j][k]表示第i个人，一个角度为j,还有一个为k的最小花费，一个人用和两个人用的情况分开讨论，然后记录dp状态转移路径。这个输出路径让这题变得麻烦了不少。只是机智的我还是把它搞♂出♂来♂了。

代码：

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

using namespace std;

#define INF 0x3f3f3f3f

#define min(a,b) ((a)<(b)?(a):(b))

const int N = 10005;

int t, n, i, j, k, dp[N][5][5], ans, an[N];

struct Stu {

	int t, l, id;

} s[N];

struct Out {

	int n, l, r, out1, out2;

} out[N][5][5];

bool cmpt(Stu a, Stu b) {

	return a.t < b.t;

}

bool cmpid(Stu a, Stu b) {

	return a.id < b.id;

}

void print(int n, int l, int r) {

	Out next = out[n][l][r];

	if (n == 0) return;

	if (next.out2 != -1) {

		an[s[n - 1].id] = next.out1;

		an[s[n].id] = next.out2;

	}

	else {

		an[s[n].id] = next.out1;

	}

	print(next.n, next.l, next.r);

}

int main() {

	scanf("%d", &t);

	while (t--) {

		ans = INF;

		memset(dp, INF, sizeof(dp));

		dp[0][0][0] = 0;

		scanf("%d", &n);

		for (i = 1; i <= n; i++) {

			scanf("%d%d", &s[i].t, &s[i].l);

			s[i].l = s[i].l / 10 - 1;

			s[i].id = i;

		}

		sort(s + 1, s + n + 1, cmpt);

		for (i = 1; i <= n; i++) {

			int tmp1 = s[i].l;

			if (i == n || s[i].t != s[i + 1].t) {

				for (j = 0; j < 5; j++) {

					for (k = 0; k < 5; k++) {

						if (dp[i][tmp1][k] > dp[i - 1][j][k] + abs(tmp1 - j) * 10) {

							dp[i][tmp1][k] = dp[i - 1][j][k] + abs(tmp1 - j) * 10;

							out[i][tmp1][k].l = j; out[i][tmp1][k].r = k; out[i][tmp1][k].n = i - 1;

							out[i][tmp1][k].out1 = 1; out[i][tmp1][k].out2 = -1;

						}

						if (dp[i][j][tmp1] > dp[i - 1][j][k] + abs(tmp1 - k) * 10) {

							dp[i][j][tmp1] = dp[i - 1][j][k] + abs(tmp1 - k) * 10;

							out[i][j][tmp1].l = j; out[i][j][tmp1].r = k; out[i][j][tmp1].n = i - 1;

							out[i][j][tmp1].out1 = 2; out[i][j][tmp1].out2 = -1;

						}

					}

				}

			}

			else {

				int tmp2 = s[i + 1].l;

				for (j = 0; j < 5; j++) {

					for (k = 0; k < 5; k++) {

						if (dp[i + 1][tmp1][tmp2] > dp[i - 1][j][k] + abs(tmp1 - j) * 10 + abs(tmp2 - k) * 10) {

							dp[i + 1][tmp1][tmp2] = dp[i - 1][j][k] + abs(tmp1 - j) * 10 + abs(tmp2 - k) * 10;

							out[i + 1][tmp1][tmp2].l = j; out[i + 1][tmp1][tmp2].r = k; out[i + 1][tmp1][tmp2].n = i - 1;

							out[i + 1][tmp1][tmp2].out1 = 1; out[i + 1][tmp1][tmp2].out2 = 2;

						}

						if (dp[i + 1][tmp2][tmp1] > dp[i - 1][j][k] + abs(tmp2 - j) * 10 + abs(tmp1 - k) * 10) {

							dp[i + 1][tmp2][tmp1] = dp[i - 1][j][k] + abs(tmp2 - j) * 10 + abs(tmp1 - k) * 10;

							out[i + 1][tmp2][tmp1].l = j; out[i + 1][tmp2][tmp1].r = k; out[i + 1][tmp2][tmp1].n = i - 1;

							out[i + 1][tmp2][tmp1].out1 = 2; out[i + 1][tmp2][tmp1].out2 = 1;

						}

					}

				}

				i++;

			}

		}

		int lv, rv;

		for (j = 0; j < 5; j++) {

			for (k = 0; k < 5; k++) {

				if (ans > dp[n][j][k] + j * 10 + k * 10) {

					ans = dp[n][j][k] + j * 10 + k * 10;

					lv = j; rv = k;

				}

			}

		}

		printf("%d\n", ans + 15 * n);

		print(n, lv, rv);

		for (i = 1; i <= n; i++)

			printf("%d\n", an[i]);

	}

	return 0;

}