2 条题解
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#include <bits/stdc++.h> using namespace std; const int N=1e5+10; vector<int>G1[N], G2[N]; int tsp, cnt, dfn[N], low[N], scc[N], num[N]; stack<int> stk;bool instk[N]; void tarjan(int x) { dfn[x]=low[x]=++tsp; stk.push(x);instk[x]=1; for(int y:G1[x]) { if(!dfn[y]) { tarjan(y); low[x]=min(low[x], low[y]); } else if(instk[y])low[x]=min(low[x], dfn[y]); } if(dfn[x]==low[x]) { cnt++; for(int z=-1;z!=x;) { z=stk.top();stk.pop();instk[z]=0; scc[z]=cnt; num[cnt]++; } } } int main() { int n, m;scanf("%d%d", &n, &m); for(int i=1, x, y; i <= m; i++) { scanf("%d%d", &x, &y); G1[x].push_back(y); } tsp=cnt=0;memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low)); memset(instk, 0, sizeof(instk)); memset(scc, 0, sizeof(scc)); memset(num, 0, sizeof(num)); for(int i=1;i <= n;i++)if(dfn[i]==0)tarjan(i); map<pair<int, int>, bool>mp; vector<int>rd(cnt+1); for(int i=1;i <= n;i++)for(int j:G1[i]) { int x=scc[i], y=scc[j]; if(x!=y && !mp[{x, y}])G2[x].push_back(y), rd[y]++, mp[{x, y}]=1; } int p=0;for(int i=1;i <= cnt;i++)if(!rd[i])p++; for(int i=1;i <= cnt;i++)if(num[i]==1 && rd[i]==0) { bool flag=0; for(int j:G2[i])if(rd[j]==1){flag=1;break;} if(!flag){p--;break;} } printf("%.6lf\n", 1.0 - 1.0*p/n ); return 0; } -
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#include<bits/stdc++.h> using namespace std; const int N=1e5+10; vector<int>G1[N],G2[N]; int tsp,cnt,dfn[N],low[N],scc[N],num[N]; stack<int> stk;bool instk[N]; void tarjan(int x) { dfn[x]=low[x]=++tsp; stk.push(x);instk[x]=1; for(int y:G1[x]) { if(!dfn[y]) { tarjan(y); low[x]=min(low[x],low[y]); } else if(instk[y])low[x]=min(low[x],dfn[y]); } if(dfn[x]==low[x]) { cnt++; for(int z=-1;z!=x;) { z=stk.top();stk.pop();instk[z]=0; scc[z]=cnt; num[cnt]++; } } } int main() { int n,m;scanf("%d%d",&n,&m); for(int i=1,x,y;i<=m;i++) { scanf("%d%d",&x,&y); G1[x].push_back(y); } tsp=cnt=0;memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low)); memset(instk,0,sizeof(instk)); memset(scc,0,sizeof(scc)); memset(num,0,sizeof(num)); for(int i=1;i<=n;i++)if(dfn[i]==0)tarjan(i); map<pair<int,int>,bool>mp; vector<int>rd(cnt+1); for(int i=1;i<=n;i++)for(int j:G1[i]) { int x=scc[i],y=scc[j]; if(x!=y && !mp[{x,y}])G2[x].push_back(y),rd[y]++,mp[{x,y}]=1; } int p=0;for(int i=1;i<=cnt;i++)if(!rd[i])p++; for(int i=1;i<=cnt;i++)if(num[i]==1 && rd[i]==0) { bool flag=0; for(int j:G2[i])if(rd[j]==1){flag=1;break;} if(!flag){p--;break;} } printf("%.6lf\n", 1.0 - 1.0*p/n ); return 0; }
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信息
- ID
- 4103
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 8
- 标签
- 递交数
- 161
- 已通过
- 22
- 上传者