首页 > 代码库 > SGU 132. Another Chocolate Maniac
SGU 132. Another Chocolate Maniac
题意:有1*2的小方块,在n*m (n <= 70, m <= 7)的棋盘内有一些点不能覆盖,要求将方块最少数量地放入棋盘,使得空格子两两不相邻。
m那么小,很明显是状态压缩DP对吧。
我们用f[i][j]表示在第i行,覆盖前一部分后当前状态为j的最小花费,转移是很显然的。重点是状态之间的关系如何确定,否则无法进行转移。考虑预处理出融洽的两个状态。用1表示在当前位放置了一块方块,我们从0枚举到(1 << m) - 1,我们要判断两个决策是否融洽,但是问题来了:棋盘上有不可覆盖点!!其实无所谓吧,我们可以尝试不预处理,而是在转移时枚举状态。复杂度:O(n*s*s)。好像没了??2维数组真的够了么???当然不行啊!因为一个方块可以同时影响两行的状态,所以我们要改进状态表示,用三维数组吧。。。然后被细节问题坑成狗。空间卡的死,需要用滚动数组优化。dfs实现时,其实只要考虑好各种情况就行了。
1 //{HEADS 2 #define FILE_IN_OUT 3 #include <cstdio> 4 #include <cstring> 5 #include <cstdlib> 6 #include <cmath> 7 #include <ctime> 8 #include <algorithm> 9 #include <iostream> 10 #include <fstream> 11 #include <vector> 12 #include <stack> 13 #include <queue> 14 #include <deque> 15 #include <map> 16 #include <set> 17 #include <bitset> 18 #include <complex> 19 #include <string> 20 #define REP(i, j) for (int i = 1; i <= j; ++i) 21 #define REPI(i, j, k) for (int i = j; i <= k; ++i) 22 #define REPD(i, j) for (int i = j; 0 < i; --i) 23 #define STLR(i, con) for (int i = 0, sz = con.size(); i < sz; ++i) 24 #define STLRD(i, con) for (int i = con.size() - 1; 0 <= i; --i) 25 #define CLR(s) memset(s, 0, sizeof s) 26 #define SET(s, v) memset(s, v, sizeof s) 27 #define mp make_pair 28 #define pb push_back 29 #define PL(k, n) for (int i = 1; i <= n; ++i) { cout << k[i] << ‘ ‘; } cout << endl 30 #define PS(k) STLR(i, k) { cout << k[i] << ‘ ‘; } cout << endl 31 using namespace std; 32 void FILE_INIT(string FILE_NAME) { 33 #ifdef FILE_IN_OUT 34 #ifndef ONLINE_JUDGE 35 freopen((FILE_NAME + ".in").c_str(), "r", stdin); 36 freopen((FILE_NAME + ".out").c_str(), "w", stdout); 37 #endif 38 #endif 39 } 40 typedef long long LL; 41 typedef double DB; 42 typedef pair<int, int> i_pair; 43 const int INF = 0x3f3f3f3f; 44 const LL INFL = 0x3f3f3f3f3f3f3f3f; 45 //} 46 47 char gchar() { 48 char ret = getchar(); 49 for(; ret != ‘.‘ && ret != ‘*‘; ret = getchar()); 50 return ret; 51 } 52 53 const int maxn = 74; 54 const int maxs = (1 << 7) + 3; 55 int G[maxn]; 56 int n, m, now, last; 57 58 LL f[2][maxs][maxs]; 59 int Snum; 60 61 //最好玩的部分开始!! 62 void dfs(int col, int pre_s, int now_s, int next_s, int cnt, LL f_last) { 63 f(0 < col && !(pre_s & (1 << (col - 1))) && !(now_s & (1 << (col - 1)))) { 64 return ; 65 } 66 if(1 < col && !(now_s & (1 << (col - 1))) && !(now_s & (1 << (col - 2)))) { 67 return ; 68 } 69 if(col == m) { 70 f[now][now_s][next_s] = min(f[now][now_s][next_s], f_last + cnt); 71 return ; 72 } 73 dfs(col + 1, pre_s, now_s, next_s, cnt, f_last); 74 if(!(now_s & (1 << (col))) && !(next_s & (1 << (col)))) { 75 dfs(col + 1, pre_s, now_s | (1 << (col)), next_s | (1 << (col)), cnt + 1, f_last); 76 } 77 if(col + 1 < m && !(now_s & (1 << col)) && !(now_s & (1 << (col + 1)))) { 78 dfs(col + 1, pre_s, now_s | (1 << (col)) | (1 << (col + 1)), next_s, cnt + 1, f_last); 79 } 80 } 81 int main() { 82 FILE_INIT("132"); 83 84 scanf("%d%d", &n, &m) 85 for(int i = 1; i <= n; ++i) { 86 for(int j = 0; j < m; ++j) { 87 char c = gchar(); 88 if(c == ‘*‘) { 89 G[i] += 1 << j; 90 } 91 } 92 } 93 Snum = (1 << m) - 1; 94 now = 0, last = 1; 95 memset(f[now], INF, sizeof f[now]); 96 f[0][Snum][G[1]] = 0; 97 for(int i = 1; i <= n; ++i) { 98 swap(now, last); 99 memset(f[now], INF, sizeof f[now]);100 for(int j = 0; j <= Snum; ++j) {101 for(int k = 0; k <= Snum; ++k) {102 if(f[last][j][k] < INFL) {103 dfs(0, j, k, G[i + 1], 0, f[last][j][k]);104 }105 }106 }107 }108 LL ans = INFL;109 for(int i = 0; i <= Snum; ++i) {110 ans = min(ans, f[now][i][0]);111 }112 printf("%lld\n", ans);113 return 0;114 }
送测试数据:
3 2....**4 4.**.*..**..*.**.7 7........................*........................8 1........1 1.3 3*.*...*.*3 7*.*.*....*...**...*.*2 7.....*.**..*.*1 7*.*.*..2 2*..*70 7..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................10 3..*..*..*..*..**..*..*..*..*..10 4*..**..**..**..**..**..**..**..**..**..*10 3..*..*..*..*..*..*..*..*..*..*10 2....................5 5.*..**......**.**.*..**..
2216301321016477774
SGU 132. Another Chocolate Maniac
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。