UVA 818 - Cutting Chains(暴力+dfs判环+位运算)
| Cutting Chains |
What a find! Anna Locke has just bought several links of chain some of which may be connected. They are made from zorkium, a material that was frequently used to manufacture jewelry in the last century, but is not used for that purpose anymore. It has its very own shine, incomparable to gold or silver, and impossible to describe to anyone who has not seen it first hand.
Anna wants the pieces joined into a single end-to-end strand of chain. She takes the links to a jeweler who tells her that the cost of joining them depends on the number of chain links that must be opened and closed. In order to minimize the cost, she carefully calculates the minimum number of links that have to be opened to rejoin all the links into a single sequence. This turns out to be more difficult than she at first thought. You must solve this problem for her.
Input
The input consists of descriptions of sets of chain links, one set per line. Each set is a list of integers delimited by one or more spaces. Every description starts with an integer n, which is the number of chain links in the set, where 1 ≤n ≤15. We will label the links 1, 2,..., n. The integers following n describe which links are connected to each other. Every connection is specified by a pair of integers i,j where 1 ≤i,j ≤n and i ≠j, indicating that chain links i and j are connected, i.e., one passes through the other. The description for each set is terminated by the pair -1 -1, which should not be processed.
The input is terminated by a description starting with n = 0. This description should not be processed and will not contain data for connected links.
Output
For each set of chain links in the input, output a single line which reads
Set N: Minimum links to open is M
where N is the set number and M is the minimal number of links that have to be opened and closed such that all links can be joined into one single chain.
| Sample Input | Output for the Sample Input |
|---|---|
5 1 2 2 3 4 5 -1 -1 7 1 2 2 3 3 1 4 5 5 6 6 7 7 4 -1 -1 4 1 2 1 3 1 4 -1 -1 3 1 2 2 3 3 1 -1 -1 3 1 2 2 1 -1 -1 0 |
Set 1: Minimum links to open is 1 Set 2: Minimum links to open is 2 Set 3: Minimum links to open is 1 Set 4: Minimum links to open is 1 Set 5: Minimum links to open is 1 |
ACM World Finals 2000, Problem C
题意：很不好理解啊。。还是参考了讨论版上面别人的算法去理解的。大概意思应该是：选几个珠子去open。然后该珠子和其他就断开了。然后拿剩下的链去以这些open的珠子为点，去连接，看能不能连成一串。。求最少的open个数。
思路：n为15.利用位运算去枚举哪几个珠子要open。然后判断剩下珠子有没有超过2个分支或者形成环，如果没有，在判断剩下的链个数有没有超过open个数-1.如果条件都符合，那么保留下最小最为答案。
代码：
#include <stdio.h>
#include <string.h>
#define min(a,b) (a)<(b)?(a):(b)
#define INF 0x3f3f3f3f
const int N = 20;
int n, g[N][N], vis[N], cn;
void init() {
int a, b;
memset(g, 0, sizeof(g));
while (~scanf("%d%d", &a, &b) && a != -1 && b != -1) {
g[a - 1][b - 1] = g[b - 1][a - 1] = 1;
}
}
bool two(int s) {
for (int i = 0; i < n; i++) {
if (s&(1<<i)) continue;
int num = 0;
for (int j = 0; j < n; j++) {
if (s&(1<<j)) continue;
if (g[i][j]) num++;
}
if (num > 2) return true;
}
return false;
}
bool dfs(int s, int now, int fa) {
vis[now] = 1;
for (int i = 0; i < n; i++) {
if (!g[now][i] || (s&(1<<i)) || i == fa) continue;
if (vis[i]) return true;
if (dfs(s, i, now)) return true;
}
return false;
}
bool circle(int s) {
for (int i = 0; i < n; i++) {
if (vis[i] || (s&(1<<i))) continue;
cn++;
if (dfs(s, i, -1)) return true;
}
return false;
}
int cal(int s) {
return s == 0 ? 0 : cal(s / 2) + (s&1);
}
int solve() {
int ans = INF;
int s = (1<<n);
for (int i = 0; i < s; i++) {
cn = 0; memset(vis, 0, sizeof(vis));
if(two(i) || circle(i)) continue;
if (cal(i) >= cn - 1)
ans = min(cal(i), ans);
}
return ans;
}
int main() {
int cas = 0;
while (~scanf("%d", &n) && n) {
init();
printf("Set %d: Minimum links to open is %d\n", ++cas, solve());
}
return 0;
}