codeforces 825E(拓扑)

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题意：要求输出字典序最小的拓扑序列。

思路：反向建个图，然后选取最大值进行拓扑。

PS :一直想不到这个反向，当初傻傻地交了一发直接图跑最小，然后回出现当选取的大的时候，会使得小的比原来的更小的情况。

#include<bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int,int> P;#define fi first#define se second#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x)#define PI acos(-1.0)#define ITER set<int>::iteratorconst int Mod = 1e9 + 7;const int maxn = 1e5 + 10;struct Edge{int to,next;}edge[maxn];int head[maxn];int edge_num;int in[maxn];int ans[maxn];void add_edge(int x,int y){edge[edge_num] = (Edge){y,head[x]};head[x] = edge_num ++;}void Init(){    clr(head,-1);edge_num = 0;clr(in,0);}priority_queue<int,vector<int>,less<int> >q;int main(){    int n,m;    while( ~ scanf("%d%d",&n,&m))    {        Init();        for(int i = 1;i <= m;i ++){int x,y;scanf("%d%d",&x,&y);add_edge(y,x);in[x] ++;}        while(!q.empty())q.pop();        for(int i = 1; i <= n;i ++){if(!in[i]){q.push(i);}}        int cnt = n;        while(!q.empty())        {            int u = q.top();q.pop();            ans[u] = cnt --;            for(int i = head[u];i != -1;i = edge[i].next)            {                int v = edge[i].to;in[v] --;                if(!in[v])q.push(v);            }        }        for(int i = 1;i <= n;i ++)printf("%d%c",ans[i],i < n ? ' ':'\n');    }    return 0;}

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