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UVaLive 7456 Least Crucial Node (并查集+暴力)
题意:求标号最小的最大割点.(删除该点后,指定点#sink能到达的点数减少最多).
析:由于不知道要去掉哪个结点,又因为只有100个结点,所以我们考虑用一个暴力,把所有的结点都去一次,然后用并查集去判断。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 100;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}int x;vector<P> v;int p[105];int Find(int x) { return x == p[x] ? x : p[x] = Find(p[x]); }int main(){ while(scanf("%d", &n) == 1 && n){ scanf("%d", &x); scanf("%d", &m); int u, vv; v.clear(); for(int i = 0; i < m; ++i){ scanf("%d %d", &u, &vv); v.push_back(P(u, vv)); } int ans = 0, cnt = 0; for(int i = 1; i <= n; ++i){ if(i == x) continue; for(int j = 1; j <= n; ++j) p[j] = j; for(int j = 0; j < v.size(); ++j){ u = v[j].first; vv = v[j].second; if(u == i || vv == i) continue; int x = Find(u); int y = Find(vv); if(x != y) p[y] = x; } map<int, int> mp; map<int, int> :: iterator it; for(int j = 1; j <= n; ++j) if(i != j) ++mp[Find(j)]; if(mp.size() <= 1) continue; int y = Find(x); if(cnt < n-mp[y]-1){ cnt = n-mp[y]-1; ans = i; } } printf("%d\n", ans); } return 0;}
UVaLive 7456 Least Crucial Node (并查集+暴力)
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