# 【luogu P1983 车站分级】 题解

step1：把题读进去，理解。得到 非停靠点的等级 < 停靠点的等级

step2：把上述不等关系转化为有向图。即由非停靠点向停靠点连一条边

step3：对于每个入度为零的点dfs找最长路。取其max

step4：输出max+1

code：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 500010;
int n, m, answer = -1e9, a[1010], start[1010], rudu[1010];
bool flag[1010];
struct edge{
int next, to, len;
}e[maxn<<2];
int dis[1010];
bool vis[1010], used[1010][1010];
void add(int u, int v, int w)
{
e[++cnt].to = v; e[cnt].next = head[u]; head[u] = cnt; e[cnt].len = w;
}
int SPFA(int s)
{
int ans = -1e9;
queue<int> q;
while(!q.empty()) q.pop();
for(int i = 1; i <= n; i++) dis[i] = -1e9;
memset(vis, 0, sizeof(vis));
dis[s] = 0, vis[s] = 1, q.push(s);
while(!q.empty())
{
int now = q.front(); q.pop();
vis[now] = 0;
for(int i = head[now]; i != -1; i = e[i].next)
{
if(dis[e[i].to] < dis[now] + e[i].len)
{
dis[e[i].to] = dis[now] + e[i].len;
if(!vis[e[i].to])
{
q.push(e[i].to);
vis[e[i].to] = 1;
}
}
}
}
for(int i = 1; i <= n; i++)
if(dis[i] != 1e9) ans = max(dis[i], ans);
return ans;
}

int main()
{
freopen("testdata.in","r",stdin);
scanf("%d%d",&n,&m);
for(int i = 1; i <= m; i++)
{
int k;
memset(flag, 0, sizeof(flag));
memset(a, 0, sizeof(a));
scanf("%d",&k);
for(int j = 1; j <= k; j++)
{
scanf("%d",&a[j]);
flag[a[j]] = 1;
}
for(int j = a[1]; j <= a[k]; j++)
{
if(flag[j] == 0)
{
for(int l = 1; l <= k; l++)
if(used[j][a[l]] == 0)
{
used[j][a[l]] = 1;
rudu[a[l]]++;
}
}
}
}//build graph
int tot = 0;
for(int i = 1; i <= n; i++)
if(rudu[i] == 0) start[++tot] = i;
for(int i = 1; i <= tot; i++)
}


posted @ 2019-06-12 20:37  Misaka_Azusa  阅读(174)  评论(0编辑  收藏
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