bzoj 4010 [HNOI 2015 Day1 T3]

听说是BestCoderBestCoder原题.

题目本意是求一个拓扑序，使得1尽量靠前，1靠前的前提下2尽量靠前。。。


我们可以反向思考，建立反图，求字典序最大的拓扑序，然后反向输出即可


（对于一个点，我们将比它大且能放它后面的点都尽量放它后面了，这样理解一下）


题解戳这里

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>

#define Eu first
#define Ev second

const int MAXN = 1e5+5, MAXM = 1e5+5;
int n, m;
std::pair<int,int> edge[MAXM];
std::priority_queue<int> heap;
bool ind[MAXN];int rdu[MAXN], cur[MAXN];;
int ans[MAXN], len;

void Init()
{
        scanf("%d%d",&n,&m);
        for(int i = 1,u,v; i <= m; i++)
        {
            scanf("%d%d",&u,&v);
            edge[i] = std::make_pair(v,u);
        }
}
void Build()
{
        std::sort(edge + 1,edge + m + 1);
        m = std::unique(edge + 1,edge + m + 1) - (edge + 1);

        memset(cur,0,sizeof(cur));
        for(int i = 1; i <= m; i++) cur[edge[i].Eu]++;
        for(int i = 1; i <= n; i++) cur[i] += cur[i-1];

        memset(rdu,0,sizeof(rdu));      
        for(int i = 1; i <= m; i++)
            rdu[edge[i].Ev] ++;

        for(int i = 1; i <= n; i++)
            if(!rdu[i]) heap.push(i); 

}

void Solve()
{
    len = 0, memset(ind,false,sizeof(ind));

    while(!heap.empty())
    {
        int t = heap.top();
        ans[++len] = t, heap.pop(); 

        for(int i = cur[t], p; i > cur[t-1]; i--)
            if(!ind[p = edge[i].Ev])
                if(!(--rdu[p])) heap.push(p), ind[p] = true;                

    }
    return; 
}
void Output()
{
    if(len == n)
    {
        for(int i = n; i >= 1; i--)
            printf("%d ",ans[i]);
        printf("\n");   
    }
    else
        puts("Impossible!");
}

int main()
{
    int T;
#ifndef ONLINE_JUDGE    
    freopen("dishes.in","r",stdin);
    freopen("dishes.out","w",stdout);
#endif  
    scanf("%d",&T);

    while(T--)
    {
        Init();
        Build();
        Solve();
        Output();
    }
#ifndef ONLINE_JUDGE    
    fclose(stdin);
    fclose(stdout);
#endif  
    return 0;
}