HDU 1824 let ' s go Home (2-sat judgment)

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Let ' s go home

Time limit:10000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1616 Accepted Submission (s): 661


Problem description As a child, homesickness is a small stamp, I am in this, mother in the head.
--Yu Guangzhong

Training is hard, the road is bumpy, rest is necessary. After a period of training, LCY decided to let everyone go home to relax, but the training is still going on, LCY came up with the following home rules, each team (three people a team) or the captain left or the remaining two players left at the same time; each pair of players, if player a leaves, then team B must go home to rest, or b leave , a go home. As the number of training team this year to break through the highest record of previous years, management difficulty is quite large, LCY also do not know whether their decision is feasible, so this problem will be handed to you, hehe, benefits ~, FREE * * Rafting day.

The first line of input has two integers, T and m,1<=t<=1000 represent the number of troops, and 1<=m<=5000 represents the logarithm.
Next there are t-lines, three integers per line, indicating the number of team members, and the first player is the team leader.
Then there are m lines, two integers per line, indicating the number of a pair of players.
Each team member belongs to only one team. The player number starts at 0.

Output Yes, otherwise output no, end with EOF.

Sample Input

1 20 1 20 11 22 40 1 23 4 50 30 41 31 4

Sample Output

Yesno

Author whisky
Sourcehdoj Summer Exercise (3)-Hold by wiskey problem-solving ideas: 2-sat is a conflict between the two even side.

#include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath > #include <vector> #include <queue> #include <algorithm> #include <stack> #include <set > #include <map> #define LL Long long using namespace std;const int maxn = 20010;vector<int>g[maxn];int pre[m AXN], LOWLINK[MAXN], SCCNO[MAXN], Dfs_clock, scc_cnt;stack<int> s;void dfs (int u) {pre[u] = lowlink[u] = ++dfs_clock ; S.push (U); int sz = G[u].size (); for (int i=0;i<sz;i++) {int v = g[u][i];if (!pre[v]) {DFS (v); Lowlink[u] = min (Lowlink[u], lowlink[v] );} else if (!sccno[v]) {Lowlink[u] = min (Lowlink[u], pre[v]);}} if (lowlink[u] = = Pre[u]) {scc_cnt++;for (;;) {int x = S.top (); S.pop (); sccno[x] = scc_cnt;if (x = = u) break;}}} void find_scc (int n) {dfs_clock = scc_cnt = 0;memset (sccno, 0, sizeof (SCCNO)), memset (pre, 0, sizeof (PRE)), for (int i=1;i&lt ; =n;i++) if (!pre[i]) DFS (i);} int N, M, T;int main () {while (scanf ("%d%d", &t, &m)!=eof) {int x, y, Z; N =3 * t;for (int i=0;i<=2*n;i++) g[i].clear (); for (int i=1;i<=t;i++) {scanf ("%d%d%d", &x, &y, &z); x + +; y++ ; z++; G[x+n].push_back (y); G[x+n].push_back (z); G[y+n].push_back (x); G[z+n].push_back (x);} for (int i=1;i<=m;i++) {scanf ("%d%d", &x, &y); x + +; y++; G[x].push_back (y + N); G[y].push_back (x + N);} FIND_SCC (2 * N); int flag = 1;for (int i=1;i<=n;i++) {if (sccno[i] = = Sccno[i + N]) {flag = 0;break;}} if (flag) puts ("yes"), Else puts ("no");} return 0;}