lightoj1037 状压DP(入门级)

题意

简单

思路

dp[sta]表示状态为sta时需要打的最少次数，dp[0] = 0;

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2016
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
#define lson rt << 1
#define rson rt << 1 | 1
#define bug cout << "BUG HERE\n"
#define debug(x) cout << #x << " = " << x << endl
#define ALL(v) (v).begin(), (v).end()
#define lowbit(x) ((x)&(-x))
#define Unique(x) sort(ALL(x)); (x).resize(unique(ALL(x)) - (x).begin())
#define BitOne(x) __builtin_popcount(x)
#define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
#define Rep(i, l, r) for (int i = l;i <= r;++i)
#define Rrep(i, r, l) for (int i = r;i >= l;--i)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-8;
const double pi = 4 * atan(1);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
int nCase = 0;
//精度正负、0的判断
int dcmp(double x){if (fabs(x) < eps) return 0;return x < 0?-1:1;}
template<class T> inline bool read(T &n){
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T> inline void write(T n){
    if(n < 0){putchar('-');n = -n;}
    int len = 0,data[20];
    while(n){data[len++] = n%10;n /= 10;}
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
LL QMOD(LL x, LL k) {
    LL res = 1LL;
    while(k) {if (k & 1) res = res * x % MOD;k >>= 1;x = x * x % MOD;}
    return res;
}
int n;
int healthy[16];
char a[16][16];
int dp[1 << 16];
// int dfs(int sta) {
//  if (!sta) return 0;
//  if (dp[sta] != -1) return dp[sta];
//  dp[sta] = inf;
//  for (int i = 0;i < n;++i) {
//      if ((sta >> i) & 1) {
//          int maxv = 1;
//          for (int j = 0;j < n;++j) {
//              maxv = max(maxv, sta >> j & 1 ? 0 : a[j][i] - '0');
//          }
//          dp[sta] = min(dp[sta], dfs(sta^(1<<i)) + (healthy[i] + maxv - 1) / maxv);
//      }
//  }
//  return dp[sta];
// }
int solve() {
    memset(dp, inf, sizeof dp);
    dp[0] = 0;
    for (int sta = 0;sta < (1 << n);++sta) {
        for (int i = 0;i < n;++i) {
            if ((sta & (1 << i)) == 0) {
                int maxv = 1;
                for (int j = 0;j < n;++j) {
                    maxv = max(maxv, sta >> j & 1 ? a[j][i] - '0' : 0);
                }
                dp[sta ^ (1 << i)] = min(dp[sta ^ (1 << i)], dp[sta] + (healthy[i] + maxv - 1) / maxv);
            }
        }
    }
    return dp[(1 << n) - 1];
}
int main(int argc, const char * argv[])
{    
    // freopen("/Users/jamesqi/Desktop/in.txt","r",stdin);
    // freopen("/Users/jamesqi/Desktop/out.txt","w",stdout);
    // ios::sync_with_stdio(false);
    // cout.sync_with_stdio(false);
    // cin.sync_with_stdio(false);

    int kase;cin >> kase;
    while(kase--) {
        cin >> n;
        Rep(i, 0, n - 1) cin >> healthy[i];
        Rep(i, 0, n - 1) cin >> a[i];
        memset(dp, -1, sizeof dp);
        int ans = solve();
        // int ans = dfs((1 << n) - 1);
        printf("Case %d: %d\n", ++nCase, ans);
    }

    // showtime;
    return 0;
}