UVA 11578

题目链接：11578 - Situp Benches

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
", ans + 15 * n);
		print(n, lv, rv);
		for (i = 1; i <= n; i++)
			printf("%d
", an[i]);
	}
	return 0;
}