UVA 11578 - Situp Benches（dp)

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，现在给定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,另一个为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。不过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;}