MicroRNA Ranking（Tehran2016）

题意：给出m个n的全排列，求一个n的全排列，满足对于i<j，至少存在一半的全排列中，ai排在aj的前面，求字典序最小方案，或者是无解。

(1)首先我们对 vis[ a[i] ][ a[j] ]++ ,求出a[i] 对 a[j] 的贡献。对vis[i][j] 和 vis[j][i]是否大于 一半 ,若大于就建立一条边，最后跑一边拓扑排序

(2)比赛的时候，将大于j 的数插入 s[j] 的集合中，然后对每一位进行筛选，若当前位置的数没有比他大的，则当前位置的数就为该数，然后将这个数从s[i] (i = 1 - > n)中删除这个数

依次循环，算的时间复杂度是0（k*n*n + n*n*log(n)），没想到超时了

比赛结束后，改成用树状数组维护，AC  ....  没想到STL 时间复杂度真的是高

拓扑排序 AC code

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 1e3 + 10;
 
int vis[N][N],a[N],in[N],ans[N],n;
 
vector<int>edge[N];
 
int topu()
{
    int cont=0;
    priority_queue<int,vector<int>,greater<int> >Q;
    for(int i=1; i<=n; i++)
        if(in[i]==0)
            Q.push(i);
    while(!Q.empty())
    {
        int u=Q.top();
        Q.pop();
        cont++;
        ans[cont] = u;
        for(int i = 0 ;i < edge[u].size(); i++)
        {
            int v = edge[u][i];
            if(--in[v]==0)
                Q.push(v);
        }
    }
    if(cont<n)
        return false;
    return true;
}
int main()
{
    int m;
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        memset(vis,0,sizeof(vis));
        memset(in,0,sizeof(in));
        for(int i = 1;i <= m;i++)
        {
            for(int j = 1;j <= n;j++)
                scanf("%d",&a[j]);
            for(int j = 1;j <= n;j++)
                for(int k = j+1;k <= n;k++)
                    vis[a[j]][a[k]]++;
        }
 
        for(int i = 1; i <= n;i++)
            for(int j = i + 1;j <= n;j++)
            {
                if(vis[i][j] > vis[j][i])   edge[i].push_back(j),in[j]++;
                else if(vis[i][j] < vis[j][i])  edge[j].push_back(i),in[i]++;
            }
 
        if(topu())
        {
            for(int i = 1;i <= n;i++)
                printf("%d%c",ans[i],i == n?'
':' ');
        }
        else
            puts("No solution");
 
        for(int i = 1;i <= n;i++)
            edge[i].clear();
 
    }
    return 0;
}

set超时代码：

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 1e3 + 10;
 
int vis[N][N],a[N],vised[N];
set<int>s[N];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        for (int i=1;i<=n;i++)
        s[i].clear();
        memset(vis,0,sizeof(vis));
        memset(vised,0,sizeof(vised));
        for(int i = 1;i <= m;i++)
        {
            for(int j = 1;j <= n;j++)
                scanf("%d",&a[j]);
            for(int j = 1;j <= n;j++)
                for(int k = j+1;k <= n;k++)
                    vis[a[j]][a[k]]++;
        }
        for(int i = 1; i <= n;i++)
            for(int j = i + 1;j <= n;j++)
            {
                if(vis[i][j] > m/2) vis[i][j] = 1,vis[j][i] = 0,s[j].insert(i);
                else if(vis[j][i] > m/2) vis[j][i] = 1,vis[i][j] = 0,s[i].insert(j);
                else    vis[i][j] = vis[j][i] = 0;
            }
 
        int ans[N],tot = 1;
 
        while(tot <= n)
        {
            int flag = 0,now = 0;
            for(int i = 1;i <= n;i++)
            {
                if(!vised[i])
                {
                    if(s[i].empty())
                    {
                        flag = 1;
                        now = i;
                        break;
                    }
                }
            }
            if(!flag)   break;
            ans[tot] = now;
            vised[now] = 1;
            for(int i = 1;i <= n;i++)
                if(s[i].find(now) != s[i].end())
                    s[i].erase(s[i].find(now));
            tot++;
        }
 
        if(tot <= n)    puts("No solution");
        else
        {
            for(int i = 1;i <= n;i++)
                printf("%d%c",ans[i],i == n?'
':' ');
        }
    }
    return 0;
}

树状数组AC code:

#include <bits/stdc++.h>
#define IT set<int>::iterator

using namespace std;

const int N = 1e3 + 10;

int vis[N][N],a[N],vised[N],c[N][N],n;


int lowbit(int x){  return x&(-x);}
void update(int pos,int x,int y)
{
    while(x <= n)
    {
        c[pos][x] += y;
        x += lowbit(x);
    }
}

int query(int pos,int x)
{
    int ans = 0;
    while(x)
    {
        ans += c[pos][x];
        x -= lowbit(x);
    }
    return ans;
}
int main()
{
    int m;
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        memset(c,0,sizeof(c));
        memset(vis,0,sizeof(vis));
        memset(vised,0,sizeof(vised));
        for(int i = 1;i <= m;i++)
        {
            for(int j = 1;j <= n;j++)
                scanf("%d",&a[j]);
            for(int j = 1;j <= n;j++)
                for(int k = j+1;k <= n;k++)
                    vis[a[j]][a[k]]++;
        }
        for(int i = 1; i <= n;i++)
            for(int j = i + 1;j <= n;j++)
            {
                if(vis[i][j] > vis[j][i])   vis[i][j] = 1,vis[j][i] = 0,update(j,i,1);
                else if(vis[j][i] > vis[i][j]) vis[j][i] = 1,vis[i][j] = 0,update(i,j,1);
                else vis[i][j] = vis[j][i] = 0;
            }

        int ans[N],tot = 1;

        while(tot <= n)
        {
            int flag = 0,now = 0;
            for(int i = 1;i <= n;i++)
            {
                if(!vised[i])
                {
                    if(!query(i,n))
                    {
                        flag = 1;
                        now = i;
                        break;
                    }
                }
            }
            if(!flag)   break;
            ans[tot] = now;
            vised[now] = 1;
            for(int i = 1;i <= n;i++)
            {
                if(vis[now][i])
                    update(i,now,-1);
            }
            tot++;
        }

        if(tot <= n)    puts("No solution");
        else
        {
            for(int i = 1;i <= n;i++)
                printf("%d%c",ans[i],i == n?'
':' ');
        }
    }
    return 0;
}