POJ 2438 (哈密顿回路)

分析:

2*n个小朋友,每个最多有n-1个"敌人",显然是存在哈密顿回路的.

预处理边,然后找哈密顿回路.

code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
#define pb push_back
#define sz(a) (int)(a).size() const int INF = 500;
bool edge[INF][INF];
typedef vector<int> vi;
vi ans;
//求哈密顿回路O(n^2)
void Hamilton (vi& ans, bool edge[INF][INF], int n) {
int s = 1, tol = 2, t, i, j;
bool vis[INF] = {0};
for (i = 1; i <= n; i++) if (edge[s][i]) break;
t = i;
vis[s] = vis[t] = 1;
ans.pb (s); ans.pb (t);
while (1) {
//头尾拓展
while (1) {
for (i = 1; i <= n; i++) {
if (edge[t][i] && !vis[i]) {
vis[i] = 1; t = i;
ans.pb (i);
break;
}
}
if (i > n) break;
}
reverse (ans.begin(), ans.end() );
swap (s, t);
while (1) {
for (i = 1; i <= n; i++) {
if (edge[t][i] && !vis[i]) {
vis[i] = 1; t = i;
ans.pb (i);
break;
}
}
if (i > n) break;
}
//如果S和T不相连
if (!edge[s][t]) {
for (i = 1; i < sz (ans) - 2; i++)
if (edge[ans[i]][t] && edge[ans[i + 1]][s]) break;
reverse (ans.begin() + i + 1, ans.end() );
t = * (ans.end() - 1);
}
tol = sz (ans);
if (tol == n) return;
//如果还有点未加入ans
for (j = 1; j <= n; j++) {
if (vis[j]) continue;
//找到与这个点相连的点
for (i = 1; i < tol - 1; i++) if (edge[ans[i]][j]) break;
if (edge[ans[i]][j]) break;
}
s = ans[i - 1], t = j;
reverse (ans.begin(), ans.begin() + i );
reverse (ans.begin() + i, ans.end() );
ans.pb (j), vis[j] = 1;
}
}
int n, m;
int main() {
while (~scanf ("%d %d", &n, &m) ) {
if (n == 0 && m == 0) return 0;
memset (edge, 1, sizeof edge);
for (int i = 1; i <= 2 * n; i++) edge[i][i] = 0;
int x, y;
for (int i = 1; i <= m; i++) {
scanf ("%d %d", &x, &y);
edge[y][x] = edge[x][y] = 0;
}
ans.clear();
Hamilton (ans, edge, n << 1);
printf ("%d", ans[0]);
for (int i = 1; i < sz (ans); i++)
printf (" %d", ans[i]);
putchar (10);
}
}

  

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