題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=3231
題意:一個三維空間內有N個立方體,立方體的邊與座標軸平行,給出一些關係,問能否滿足並構造出解,關係有兩種:
1.某兩個立方體相交
2.某個立方體所有點某一維座標 < 另一個立方體所有點該維座標
思路:考慮每一維座標,設兩個立方體左右邊界非別爲L1,R1,L2,R2,則有L1 < R1,L2 < R2,對於第一個條件,則相當於一個立方體的某一面嵌入到了另一個立方體裏面,則有L1 < R2 && L2 < R1,對於第二個條件則有R1 < L2,同理在另外兩維上也需要滿足這樣的關係,故而可以對於每一維使用拓撲排序,如果某一維關係出現矛盾就是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);
}
}
else
addedge(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;
}