HDU 1815， POJ 2749 Building roads

题目链接HDU
题目链接POJ

题意:
有n个牛棚， 还有两个中转站S1和S2， S1和S2用一条路连接起来。

为了使得随意牛棚两个都能够有道路联通，如今要让每一个牛棚都连接一条路到S1或者S2。

有a对牛棚互相有仇恨，所以不能让他们的路连接到同一个中转站。

还有b对牛棚互相喜欢，所以他们的路必须连到同一个中专站。

道路的长度是两点的曼哈顿距离。
问最小的随意两牛棚间的距离中的最大值是多少？

思路：二分距离。考虑每两个牛棚之间4种连边方式，然后依据二分的长度建立表达式，然后跑2-sat推断就可以

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;const int MAXNODE = 505;struct TwoSet {int n;vector<int> g[MAXNODE * 2];bool mark[MAXNODE * 2];int S[MAXNODE * 2], sn;void init(int tot) {n = tot * 2;for (int i = 0; i < n; i += 2) {g[i].clear();g[i^1].clear();}memset(mark, false, sizeof(mark));}void add_Edge(int u, int uval, int v, int vval) {u = u * 2 + uval;v = v * 2 + vval;g[u^1].push_back(v);g[v^1].push_back(u);}void delete_Edge(int u, int uval, int v, int vval) {u = u * 2 + uval;v = v * 2 + vval;g[u^1].pop_back();g[v^1].pop_back();}bool dfs(int u) {if (mark[u^1]) return false;if (mark[u]) return true;mark[u] = true;S[sn++] = u;for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (!dfs(v)) return false;}return true;}bool solve() {for (int i = 0; i < n; i += 2) {if (!mark[i] && !mark[i + 1]) {sn = 0;if (!dfs(i)){for (int j = 0; j < sn; j++)mark[S[j]] = false;sn = 0;if (!dfs(i + 1)) return false;}}}return true;}
} gao;const int N = 505;
int n, a, b;struct Point {int x, y;void read() {scanf("%d%d", &x, &y);}
} s1, s2, p[N], A[N * 2], B[N * 2];int dis(Point a, Point b) {int dx = a.x - b.x;int dy = a.y - b.y;return abs(dx) + abs(dy);
}int g[N][N][4];bool judge(int d) {gao.init(n);for (int i = 0; i < a; i++) {gao.add_Edge(A[i].x - 1, 0, A[i].y - 1, 0);gao.add_Edge(A[i].x - 1, 1, A[i].y - 1, 1);}for (int i = 0; i < b; i++) {gao.add_Edge(B[i].x - 1, 0, B[i].x - 1, 1);gao.add_Edge(B[i].x - 1, 0, B[i].y - 1, 1);gao.add_Edge(B[i].y - 1, 0, B[i].x -1 , 1);gao.add_Edge(B[i].y - 1, 0, B[i].y - 1, 1);}for (int i = 0; i < n; i++) {for (int j = 0; j < i; j++) {if (g[i][j][3] > d)gao.add_Edge(i, 0, j, 1);if (g[i][j][2] > d)gao.add_Edge(i, 1, j, 0);if (g[i][j][1] > d)gao.add_Edge(i, 0, j, 0);if (g[i][j][0] > d)gao.add_Edge(i, 1, j, 1);}}return gao.solve();
}int main() {while (~scanf("%d%d%d", &n, &a, &b)) {s1.read(); s2.read();for (int i = 0; i < n; i++) {p[i].read();for (int j = 0; j < i; j++) {g[i][j][0] = dis(p[i], s1) + dis(p[j], s1);g[i][j][1] = dis(p[i], s2) + dis(p[j], s2);g[i][j][2] = dis(p[i], s1) + dis(p[j], s2) + dis(s1, s2);g[i][j][3] = dis(p[i], s2) + dis(p[j], s1) + dis(s1, s2);}}for (int i = 0; i < a; i++) A[i].read();for (int i = 0; i < b; i++) B[i].read();int l = 0, r = 7777777;if (!judge(r)) printf("-1\n");else {while (l < r) {int mid = (l + r) / 2;if (judge(mid)) r = mid;else l = mid + 1;}printf("%d\n", l);}}return 0;
}