Description
A test for allergy is conducted over the course of several days, and consists of exposing you to different substances (so called allergens). The goal is to decide exactly which of the allergens you are allergic to. Each allergen has a live duration D measured in whole days, indicating exactly how many days you will suffer from an allergic reaction if you are allergic to that particular substance. An allergic reaction starts to show almost immediately after you have been exposed to an allergen which you are allergic to. The test scheme has two action points per day:
The first line of the input file contains an integer N (N<30) which denotes the total number of test cases. The description of each test case is given below:
The first line of the input contains a single integer k (1 ≤ k ≤ 20) specifying the number of allergens being tested for. Then follow k lines each containing an integer D (1 ≤ D ≤ 7) specifying the live duration of each allergen.
For each test case, print in a single line the number of days of the shortest conclusive non-adaptive test scheme. A scheme ends the morning when you no longer have active allergens in your body, thus a test scheme for a single allergen with live duration D takes D days.
2 3 2 2 2 5 1 4 2 5 2
5 10
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <map> 5 #include <string> 6 #include <cmath> 7 #include <iostream> 8 #include <time.h> 9 #include <algorithm> 10 using namespace std; 11 #define ll long long 12 #define INF 0x7fffffff 13 #define mod 1000000000 14 #define maxn (1<<20) 15 int n, m, k, s, t, ans; 16 int dp[maxn][9]; 17 int a[22],in[22],vis[22]; 18 void dfs(int *b, int t){ 19 int f[22]; 20 if (t >= ans)return; 21 int flag = 1; 22 for (int i = 0; i < n; i++){ 23 if (in[i] == 0)flag = 0; 24 if (b[i]<a[i])flag = 0; 25 } 26 if (flag){ 27 ans = min(ans, t); return; 28 } 29 for (int i = 0; i < n; i++){ 30 if (vis[i] == 1)b[i]++; 31 } 32 dfs(b, t + 1); 33 for (int i = 0; i < n; i++){ 34 if (vis[i] == 1)b[i]--; 35 } 36 int x = 0,p; 37 for (int i = 0; i < n; i++){ 38 if ((vis[i] && b[i] < a[i])){ 39 x++; 40 p = i; 41 } 42 } 43 if (x == 1&&in[p]==0){ 44 in[p] = 1; 45 dfs(b, t); 46 in[p] = 0; 47 } 48 for (int i = 0; i < n;i++) 49 if (vis[i] == 1)b[i]++; 50 for (int i = 0; i < n; i++){ 51 if (b[i]>=a[i]||vis[i])continue; 52 vis[i] = 1; 53 dfs(b, t + 1); 54 vis[i] = 0; 55 } 56 for (int i = 0; i < n; i++) 57 if (vis[i] == 1)b[i]--; 58 } 59 int main(){ 60 int T; scanf("%d", &T); 61 while (T--){ 62 int b[22]; 63 scanf("%d", &n); 64 for (int i = 0; i < n; i++){ 65 scanf("%d", &a[i]); 66 } 67 ans = 82; 68 for (int i = 0; i < n; i++){ 69 memset(in, 0, sizeof in); 70 memset(vis, 0, sizeof vis); 71 memset(b, 0, sizeof b); 72 vis[i] = 1; 73 dfs(b, 0); 74 } 75 printf("%d\n", ans); 76 } 77 return 0; 78 }
TLE方法。。。dfs这个dfs也是写的要死。。
原文：http://www.cnblogs.com/HaibaraAi/p/3854942.html
