HDU 3231 Box Relations（拓扑排序）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3231

题意：一个三维空间内有N个立方体，立方体的边与坐标轴平行，给出一些关系，问能否满足并构造出解，关系有两种：

1.某两个立方体相交

2.某个立方体所有点某一维坐标 < 另一个立方体所有点该维坐标

思路：考虑每一维坐标，设两个立方体左右边界非别为L1，R1，L2，R2，则有L1 < R1，L2 < R2，对于第一个条件，则相当于一个立方体的某一面嵌入到了另一个立方体里面，则有L1 < R2 && L2 < R1，对于第二个条件则有R1 < L2，同理在另外两维上也需要满足这样的关系，故而可以对于每一维使用拓扑排序，如果某一维关系出现矛盾就是IMPOSSIBLE

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <utility>#include <cmath>#include <queue>#include <climits>#include <functional>#include <deque>#include <ctime>#include <string>#include <set>#include <map>#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn = 3000;const int maxm = 500100;int num;int cnt[3];int head[3][maxn];int dep[3][maxn], in[3][maxn];struct edge{int to, next;} e[3][maxm];void init(){memset(cnt, 0, sizeof(cnt));memset(head, -1, sizeof(head));}void addedge(int u, int v, int id){e[id][cnt[id]].to = v;e[id][cnt[id]].next = head[id][u];head[id][u] = cnt[id]++;in[id][v]++;}bool toposort(int id){int tot = 0;queue <int> que;for (int i = 1; i <= num; i++)if (in[id][i] == 0){que.push(i);tot++;}while (!que.empty()){int u = que.front();que.pop();for (int i = head[id][u]; ~i; i = e[id][i].next){int v = e[id][i].to;--in[id][v];if (!in[id][v]){tot++;que.push(v);dep[id][v] = dep[id][u] + 1;}}}return tot == num;}int main(){int n, m;int ca = 1;while (~scanf("%d%d", &n, &m) && (n + m)){init();memset(dep, 0, sizeof(dep));memset(in, 0, sizeof(in));num = n * 2;for (int i = 0; i < 3; i++)for (int j = 1; j <= n; j++)addedge(j, j + n, i);for (int i = 0; i < m; i++){char s[5];int u, v;scanf("%s%d%d", s, &u, &v);if (s[0] == 'I'){for (int i = 0; i < 3; i++){addedge(u, v + n, i);addedge(v, u + n, i);}}elseaddedge(u + n, v, s[0] - 'X');}printf("Case %d: ", ca++);if (!toposort(0) || !toposort(1) || !toposort(2))printf("IMPOSSIBLE\n\n");else{printf("POSSIBLE\n");for (int i = 1; i <= n; i++)printf("%d %d %d %d %d %d\n", dep[0][i], dep[1][i], dep[2][i], dep[0][i + n], dep[1][i + n], dep[2][i + n]);printf("\n");}}return 0;}

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