http://www.lydsy.com/JudgeOnline/problem.php?id=3887||
https://www.luogu.org/problem/show?pid=3119
Description
In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way cow path connecting a pair of fields. For example, if a path connects from field X to field Y, then cows are allowed to travel from X to Y but not from Y to X. Bessie the cow, as we all know, enjoys eating grass from as many fields as possible. She always starts in field 1 at the beginning of the day and visits a sequence of fields, returning to field 1 at the end of the day. She tries to maximize the number of distinct fields along her route, since she gets to eat the grass in each one (if she visits a field multiple times, she only eats the grass there once). As one might imagine, Bessie is not particularly happy about the one-way restriction on FJ's paths, since this will likely reduce the number of distinct fields she can possibly visit along her daily route. She wonders how much grass she will be able to eat if she breaks the rules and follows up to one path in the wrong direction. Please compute the maximum number of distinct fields she can visit along a route starting and ending at field 1, where she can follow up to one path along the route in the wrong direction. Bessie can only travel backwards at most once in her journey. In particular, she cannot even take the same path backwards twice.
给一个有向图,然后选一条路径起点终点都为1的路径出来,有一次机会可以沿某条边逆方向走,问最多有多少个点可以被经过?(一个点在路径中无论出现多少正整数次对答案的贡献均为1)
Input
Output
Sample Input
1 2
3 1
2 5
2 4
3 7
3 5
3 6
6 5
7 2
4 7
Sample Output
HINT
Source
先把原图缩点,跑出从1到n和从n到1的最多可以遍历的牧场数,
枚举每个边做无向边的情况更新答案
SPFA跑最多牧场数
1 #include <cstdio> 2 #include <queue> 3 #define min(a,b) (a<b?a:b) 4 #define max(a,b) (a>b?a:b) 5 6 const int INF(0x3f3f3f3f); 7 const int N(1e5+5); 8 int n,head[N],sumedge; 9 struct Edge 10 { 11 int v,next; 12 Edge(int v=0,int next=0):v(v),next(next){} 13 }edge[N]; 14 inline void ins(int u,int v) 15 { 16 edge[++sumedge]=Edge(v,head[u]); 17 head[u]=sumedge; 18 } 19 20 int tim,dfn[N],low[N]; 21 int top,Stack[N],instack[N]; 22 int sumcol,col[N],point[N]; 23 void DFS(int u) 24 { 25 low[u]=dfn[u]=++tim; 26 Stack[++top]=u; instack[u]=1; 27 for(int v,i=head[u];i;i=edge[i].next) 28 { 29 v=edge[i].v; 30 if(!dfn[v]) DFS(v),low[u]=min(low[u],low[v]); 31 else if(instack[v]) low[u]=min(low[u],dfn[v]); 32 } 33 if(low[u]==dfn[u]) 34 { 35 col[u]=++sumcol; 36 point[sumcol]++; 37 for(;Stack[top]!=u;top--) 38 { 39 point[sumcol]++; 40 col[Stack[top]]=sumcol; 41 instack[Stack[top]]=0; 42 } 43 instack[u]=0; top--; 44 } 45 } 46 47 int hed[N],had[N],sum; 48 struct E 49 { 50 int v,next,w; 51 E(int v=0,int next=0,int w=0):v(v),next(next),w(w){} 52 }e[N][2]; 53 inline void insert(int u,int v) 54 { 55 e[++sum][0]=E(v,hed[u],point[v]); 56 hed[u]=sum; 57 e[sum][1]=E(u,had[v],point[u]); 58 had[v]=sum; 59 } 60 61 bool inq[N]; 62 int v1[N],v2[N]; 63 void SPFA(int op,int s,int *val,int *head) 64 { 65 for(int i=1;i<=sumcol;i++) 66 inq[i]=0,val[i]=-INF; 67 val[s]=point[s]; 68 std::queue<int>que; 69 que.push(s); 70 for(int u,v;!que.empty();) 71 { 72 u=que.front(); que.pop(); inq[u]=0; 73 for(int i=head[u];i;i=e[i][op].next) 74 { 75 v=e[i][op].v; 76 if(val[v]<val[u]+e[i][op].w) 77 { 78 val[v]=val[u]+e[i][op].w; 79 if(!inq[v]) inq[v]++,que.push(v); 80 } 81 } 82 } 83 } 84 85 inline void read(int &x) 86 { 87 x=0; register char ch=getchar(); 88 for(;ch>'9'||ch<'0';) ch=getchar(); 89 for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; 90 } 91 92 int AC() 93 { 94 int m; read(n),read(m); 95 for(int u,v;m--;) 96 read(u),read(v),ins(u,v); 97 for(int i=1;i<=n;i++) 98 if(!dfn[i]) DFS(i); 99 for(int v,u=1;u<=n;u++) 100 for(int i=head[u];i;i=edge[i].next) 101 { 102 v=edge[i].v; 103 if(col[u]!=col[v]) insert(col[u],col[v]); 104 } 105 int ans=-INF; 106 SPFA(0,col[1],v1,hed); 107 SPFA(1,col[1],v2,had); 108 for(int v,u=1;u<=sum;u++) 109 for(int i=hed[u];i;i=e[i][0].next) 110 { 111 v=e[i][0].v; 112 ans=max(ans,v1[v]+v2[u]); 113 } 114 for(int v,u=1;u<=sum;u++) 115 for(int i=had[u];i;i=e[i][1].next) 116 { 117 v=e[i][1].v; 118 ans=max(ans,v1[u]+v2[v]); 119 } 120 printf("%d ",ans-point[col[1]]); 121 return 0; 122 } 123 124 int Hope=AC(); 125 int main(){;}
Topsort跑最多牧场数
1 #include <cstdio> 2 #include <queue> 3 #define min(a,b) (a<b?a:b) 4 #define max(a,b) (a>b?a:b) 5 6 const int INF(0x3f3f3f3f); 7 const int N(1e5+5); 8 int n,head[N],sumedge; 9 struct Edge 10 { 11 int v,next; 12 Edge(int v=0,int next=0):v(v),next(next){} 13 }edge[N]; 14 inline void ins(int u,int v) 15 { 16 edge[++sumedge]=Edge(v,head[u]); 17 head[u]=sumedge; 18 } 19 20 int tim,dfn[N],low[N]; 21 int top,Stack[N],instack[N]; 22 int sumcol,col[N],point[N],rd[N],cd[N]; 23 void DFS(int u) 24 { 25 low[u]=dfn[u]=++tim; 26 Stack[++top]=u; instack[u]=1; 27 for(int v,i=head[u];i;i=edge[i].next) 28 { 29 v=edge[i].v; 30 if(!dfn[v]) DFS(v),low[u]=min(low[u],low[v]); 31 else if(instack[v]) low[u]=min(low[u],dfn[v]); 32 } 33 if(low[u]==dfn[u]) 34 { 35 col[u]=++sumcol; 36 point[sumcol]++; 37 for(;Stack[top]!=u;top--) 38 { 39 point[sumcol]++; 40 col[Stack[top]]=sumcol; 41 instack[Stack[top]]=0; 42 } 43 instack[u]=0; top--; 44 } 45 } 46 47 int hed[N],had[N],sum; 48 struct E 49 { 50 int v,next,w; 51 E(int v=0,int next=0,int w=0):v(v),next(next),w(w){} 52 }e[N][2]; 53 inline void insert(int u,int v) 54 { 55 e[++sum][0]=E(v,hed[u],point[v]); 56 hed[u]=sum; 57 e[sum][1]=E(u,had[v],point[u]); 58 had[v]=sum; 59 } 60 61 int v1[N],v2[N]; 62 #define max(a,b) (a>b?a:b) 63 void Topsort(int op,int s,int *val,int *head,int *du) 64 { 65 std::queue<int>que; 66 for(int i=1;i<=sumcol;i++) 67 { 68 if(!du[i]) que.push(i); 69 val[i]=-INF; 70 } 71 val[s]=point[s]; 72 for(int u,v;!que.empty();) 73 { 74 u=que.front(); que.pop(); 75 for(int i=head[u];i;i=e[i][op].next) 76 { 77 v=e[i][op].v; 78 val[v]=max(val[v],val[u]+e[i][op].w); 79 if(--du[v]==0) que.push(v); 80 } 81 } 82 } 83 84 inline void read(int &x) 85 { 86 x=0; register char ch=getchar(); 87 for(;ch>'9'||ch<'0';) ch=getchar(); 88 for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; 89 } 90 91 int AC() 92 { 93 int m; read(n),read(m); 94 for(int u,v;m--;) 95 read(u),read(v),ins(u,v); 96 for(int i=1;i<=n;i++) 97 if(!dfn[i]) DFS(i); 98 for(int v,u=1;u<=n;u++) 99 for(int i=head[u];i;i=edge[i].next) 100 { 101 v=edge[i].v; 102 if(col[u]==col[v]) continue; 103 rd[col[v]]++,cd[col[u]]++; 104 insert(col[u],col[v]); 105 } 106 int ans=-INF; 107 Topsort(0,col[1],v1,hed,rd); 108 Topsort(1,col[1],v2,had,cd); 109 for(int v,u=1;u<=sum;u++) 110 for(int i=hed[u];i;i=e[i][0].next) 111 { 112 v=e[i][0].v; 113 ans=max(ans,v1[v]+v2[u]); 114 } 115 for(int v,u=1;u<=sum;u++) 116 for(int i=had[u];i;i=e[i][1].next) 117 { 118 v=e[i][1].v; 119 ans=max(ans,v1[u]+v2[v]); 120 } 121 printf("%d ",ans-point[col[1]]); 122 return 0; 123 } 124 125 int Hope=AC(); 126 int main(){;}