标签：pair 多少 cto sig amp comm 面积 bsp string
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)
标签：pair 多少 cto sig amp comm 面积 bsp string
原文地址：http://www.cnblogs.com/dwtfukgv/p/6947413.html
