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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; }
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