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POJ 2836 Rectangular Covering (状压DP)
题意:平面上有 n (2 ≤ n ≤ 15) 个点,现用平行于坐标轴的矩形去覆盖所有点,每个矩形至少盖两个点,矩形面积不可为0,求这些矩形的最小面积。
析:先预处理所有的矩形,然后dp[s] 表示 状态 s 时,最少需要的面积是多少。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 100000 + 10; const int mod = 100000000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int x[20], y[20]; struct Node{ int area, cover; Node(int s,int c) : area(s), cover(c) { } }; vector<Node> rec; void calc(int i, int j, int &s, int &cover){ int w = max(abs(x[i] - x[j]), 1); int l = max(abs(y[i] - y[j]), 1); s = w * l; cover = 0; int minx = min(x[i], x[j]); int maxx = max(x[j], x[i]); int miny = min(y[i], y[j]); int maxy = max(y[j], y[i]); for(int i = 0; i < n; ++i) if(x[i] >= minx && y[i] <= maxy && x[i] <= maxx && y[i] >= miny) cover |= 1<<i; } int dp[1<<15]; int main(){ while(scanf("%d", &n) == 1 && n){ rec.clear(); for(int i = 0; i < n; ++i) scanf("%d %d", x+i, y+i); for(int i = 1; i < n; ++i) for(int j = 0; j < i; ++j){ int cover, s; calc(i, j, s, cover); rec.push_back(Node(s, cover)); } memset(dp, INF, sizeof dp); dp[0] = 0; int all = 1 << n; for(int j = 0; j < rec.size(); ++j){ Node &u = rec[j]; for(int i = 0; i < all; ++i){ if(dp[i] == INF) continue; dp[i|u.cover] = min(dp[i|u.cover], dp[i] + u.area); } } printf("%d\n", dp[all-1]); } return 0; }
POJ 2836 Rectangular Covering (状压DP)
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