usaco Controlling Companies
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首先,根据题目,只有A对B的控制超过50%,A才有可能通过B公司控制其他公司,所以我们就可以对dfs进行极大的剪枝
如果A控制B公司,那就对所有的公司,对A加上B控制的股份,
code
/* ID:yueqi LANG:C++ TASK:concom*/#include <set>#include <map>#include <ctime>#include <queue>#include <cmath>#include <stack>#include <limits.h>#include <vector>#include <bitset>#include <string>#include <cstdio>#include <cstring>#include <fstream>#include <string.h>#include <iostream>#include <algorithm>#define Si set<int>#define LL long long#define pb push_back#define PS printf(" ")#define Vi vector<int>#define LN printf("\n")#define lson l,m,rt << 1#define rson m+1,r,rt<<1|1#define SD(a) scanf("%d",&a)#define PD(a) printf("%d",a)#define SET(a,b) memset(a,b,sizeof(a))#define FF(i,a) for(int i(0);i<(a);i++)#define FD(i,a) for(int i(a);i>=(1);i--)#define FOR(i,a,b) for(int i(a);i<=(b);i++)#define FOD(i,a,b) for(int i(a);i>=(b);i--)#define readf freopen("concom.in","r",stdin)#define writef freopen("concom.out","w",stdout)const int maxn = 101;const long long BigP=999983;const long long INF = 0x5fffffff;const int dx[]={0,1,0,-1};const int dy[]={1,0,-1,0};const double pi = acos(-1.0);const double eps= 1e-7;using namespace std;int N,a[maxn][maxn];int ans[maxn][maxn],top;int dfs(int i,int j){ FOR(k,1,top){ if(i!=k && ans[i][k]<50){ ans[i][k]+=a[j][k]; if(ans[i][k]>=50) dfs(i,k); } }}int main(){ readf; writef; int p,q,c; SD(N); FOR(i,1,N){ SD(p);SD(q);SD(c); a[p][q]=c; ans[p][q]=c; if(p>top) top=p; if(q>top) top=q; } FOR(i,1,top) FOR(j,1,top){ if(ans[i][j]>=50) dfs(i,j); } FOR(i,1,top) FOR(j,1,top){ if(i!=j && ans[i][j]>=50){ PD(i);PS;PD(j);LN; } } return 0;}
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