題目描述

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.

約翰有n塊草場，編號1到n，這些草場由若幹條單行道相連。奶牛貝西是美味牧草的鑒賞家，她想到達盡可能多的草場去品嘗牧草。

貝西總是從1號草場出發，最後回到1號草場。她想經過盡可能多的草場，貝西在通一個草場只吃一次草，所以一個草場可以經過多次。因為草場是單行道連接，這給貝西的品鑒工作帶來了很大的不便，貝西想偷偷逆向行走一次，但最多只能有一次逆行。問，貝西最多能吃到多少個草場的牧草。

輸入輸出格式

輸入格式：

INPUT: (file grass.in)

The first line of input contains N and M, giving the number of fields and the number of one-way paths (1 <= N, M <= 100,000).

The following M lines each describe a one-way cow path. Each line contains two distinct field numbers X and Y, corresponding to a cow path from X to Y. The same cow path will never appear more than once.

輸出格式：

OUTPUT: (file grass.out)

A single line indicating the maximum number of distinct fields Bessie

can visit along a route starting and ending at field 1, given that she can

follow at most one path along this route in the wrong direction.

輸入輸出樣例

7 10 
1 2 
3 1 
2 5 
2 4 
3 7 
3 5 
3 6 
6 5 
7 2 
4 7 

6 

說明

SOLUTION NOTES:

Here is an ASCII drawing of the sample input:

v---3-->6

7 |\ |

^\ v \ |

| \ 1 | | | v | v 5

4<--2---^

Bessie can visit pastures 1, 2, 4, 7, 2, 5, 3, 1 by traveling

backwards on the path between 5 and 3. When she arrives at 3 she

cannot reach 6 without following another backwards path.

1A爽，縮點後變成一棵樹，spfa處理出樹上每個點到1所在點的最長路，然後枚舉連向點1和點1連向的點之間的邊，答案就是這兩個點最長路的和

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100007;
inline int read() {
    int x=0,f=1;
    char c=getchar();
    while(c<‘0‘||c>‘9‘) {
        if(c==‘-‘)f=-1;
        c=getchar();
    }
    while(c<=‘9‘&&c>=‘0‘) {
        x=x*10+c-‘0‘;
        c=getchar();
    }
    return x*f;
}
int n,m;
struct node{
    int v,next;
}edge[maxn],E1[maxn],E2[maxn];
int head[maxn],H1[maxn],H2[maxn],num,N1,N2;
void add_edge(int x,int y) {
    edge[++num].v=y;edge[num].next=head[x];head[x]=num;
}
void A1(int x,int y) {
    E1[++N1].v=y;E1[N1].next=H1[x];H1[x]=N1;
}
void A2(int x,int y){
    E2[++N2].v=y;E2[N2].next=H2[x];H2[x]=N2;
}
int dfn[maxn],low[maxn],tn=0,stack[maxn],color[maxn],color_cnt;
int top=0,cnt[maxn];bool vis[maxn];
void tarjan(int x) {
    dfn[x]=low[x]=++tn;stack[++top]=x,vis[x]=1;
    for(int i=head[x];i;i=edge[i].next) {
        int v=edge[i].v;
        if(!dfn[v]) {
            tarjan(v);
            low[x]=min(low[x],low[v]);
        }
        else if(vis[v]) low[x]=min(low[x],dfn[v]);
    }
    if(dfn[x]==low[x]) {
        ++color_cnt;
        while(stack[top]!=x) {
            vis[stack[top]]=0;
            color[stack[top]]=color_cnt;
            top--;
            cnt[color_cnt]++;
        }
        cnt[color_cnt]++;
        color[x]=color_cnt;vis[x]=0;top--;
    }
}
int dis1[maxn],dis2[maxn];
int que[maxn*4];
bool can[maxn],can2[maxn];
void topo1(int S) {
    int h=1,t=1;
    que[1]=S;
    dis1[S]=cnt[S];
    memset(vis,0,sizeof vis);
    vis[S]=1;can[S]=1;
    while(h<=t) {
        int u=que[h++];
        for(int i=H1[u];i;i=E1[i].next) {
            int v=E1[i].v;can[v]=1;
            if(dis1[v]<dis1[u]+cnt[v]) {
                dis1[v]=max(dis1[v],dis1[u]+cnt[v]);
                if(!vis[v])que[++t]=v,vis[v]=1;
            }
        }
        vis[u]=0;
    }
}
void topo2(int S) {
    int h=1,t=1;
    que[1]=S;
    dis2[S]=cnt[S];    
    memset(vis,0,sizeof vis);
    vis[S]=1;can2[S]=1;
    while(h<=t) {
        int u=que[h++];
        for(int i=H2[u];i;i=E2[i].next) {
            int v=E2[i].v;can2[v]=1;
            if(dis2[v]<dis2[u]+cnt[v]) {
                dis2[v]=max(dis2[v],dis2[u]+cnt[v]);
                if(!vis[v])que[++t]=v,vis[v]=1;
            }
        }
        vis[u]=0;
    }
}
int main() {
    n=read(),m=read();
    for(int a,b,i=1;i<=m;++i) {
        a=read(),b=read();
        add_edge(a,b);
    }
    for(int i=1;i<=n;++i) 
        if(!dfn[i])tarjan(i);
    int S=color[1];
    for(int i=1;i<=n;++i) {
        for(int j=head[i];j;j=edge[j].next) {
            int v=edge[j].v;
            if(color[v]!=color[i]){
                A1(color[i],color[v]);
                A2(color[v],color[i]);
            }
        }
    }
    topo1(color[1]);
    topo2(color[1]);
    int ans=0;
    for(int i=1;i<=color_cnt;++i) {
        for(int j=H1[i];j;j=E1[j].next) {
            int v=E1[j].v;
            if(can2[i]&&can[v]) {
                ans=max(ans,dis2[i]+dis1[v]-cnt[color[1]]);
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}

luogu P3119 [USACO15JAN]草鑒定Grass Cownoisseur