poj2836 状态压缩dp

来源：互联网发布：有关网络的英语单词编辑：程序博客网时间：2024/05/22 01:04

分析：

两个点形成矩形可定时连接对角线的顶点。
预处理所有可能矩形的面积和他们能覆盖的点O(n3)。
dp[s]表示当前覆盖情况的最小面积和，枚举加入每个矩形即可O(n22n)。
注意两点连线平行坐标轴的情况和矩形顶点必须是整数这个条件。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;struct jibancanyang{    int dp[1 << 15], n;    int matrixs[15 * 15][2], points[15][2], cnt;    void preProcess() {        for (int i = 0; i < n; ++i) {            for (int j = i + 1; j < n; ++j) {                int a = abs(points[i][0] - points[j][0]), b = abs(points[i][1] - points[j][1]);                if (a == 0) a = 1;                if (b == 0) b = 1;                matrixs[cnt][0] = a * b;                matrixs[cnt][1] = 0;                for (int k = 0; k < n; ++k) {                    if (min(points[i][0], points[j][0]) <= points[k][0] && points[k][0] <= max(points[i][0], points[j][0])                        && min(points[i][1], points[j][1]) <= points[k][1] && points[k][1] <= max(points[i][1], points[j][1]))                            matrixs[cnt][1] |= 1 << k;                }                cnt++;            }        }    }    int dynamicProgramming() {        memset(dp, 0x3f, sizeof(dp));        dp[0] = 0;        for (int s = 0; s < 1 << n; ++s) {            for (int i = 0; i < cnt; ++i) {                dp[s | matrixs[i][1]] = min(dp[s | matrixs[i][1]], dp[s] + matrixs[i][0]);            }        }        return dp[(1 << n) - 1];    }    void fun() {        while (scanf("%d", &n), n) {            cnt = 0;            for (int i = 0; i < n; ++i) {                scanf("%d%d", &points[i][0], &points[i][1]);            }            preProcess();            printf("%d\n", dynamicProgramming());        }    }}ac;int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    ac.fun();    return 0;}

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