[BZOJ5407]girls

也是CF985G。。。

容斥+三元环计数

CF数据太弱啦

vis没赋初值-1竟然过了QAQ

所以又调了我半个小时才搞掉QAQ

数数真难QAQ

记得要写#include<vector>！！！

Dev给加的奇奇怪怪的编译选项会给你自动填补的！！！

QAQ长个记性QAQ

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<vector>
#define ull unsigned long long
#define ll long long
#define inf 20021225
#define N 400010
#define it vector<int>::iterator
using namespace std;
 
int sz[N],deg[N],n,m;
vector<ull> sum[N];
vector<int> ed[N];
struct edge{int to,lt;}e[N];
int in[N],cnt; int vis[N];
void add(int x,int y){e[++cnt].to=y;e[cnt].lt=in[x];in[x]=cnt;}
void go(int x){for(int i=in[x];i;i=e[i].lt) vis[e[i].to]=x;}
ull calc(int x){if(x<0)  return 0; return (ll)x*(x-1)/2;}
ull up(int l,int r){if(l>r)  return 0; return (ll)(r+l)*(r-l+1)/2;}
int main()
{
    scanf("%d%d",&n,&m); ull A,B,C,res=0;
    scanf("%llu%llu%llu",&A,&B,&C); int x,y;
    //scanf("%I64d%I64d%I64d",&A,&B,&C); int x,y;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        ed[x].push_back(y);
        ed[y].push_back(x);
    }
    //ull edg=(ull)n*(n-1)>>1;
    //at least 0
    for(int i=0;i<n;i++)
        res+=A*i*calc(n-i-1),
        res+=B*i*i*(n-i-1),
        res+=C*i*calc(i),
        sort(ed[i].begin(),ed[i].end()),
        sum[i].push_back(0);
    //printf("%llu\n",res);
    for(x=0;x<n;x++) for(it i=ed[x].begin();i!=ed[x].end();i++)
        sz[x]=(*i)>x?sz[x]:sz[x]+1,++deg[x],sum[x].push_back(sum[x][deg[x]-1]+(*i));
    //at least 1
    for(x=0;x<n;x++) for(it i=ed[x].begin();i!=ed[x].end();i++)
        if((*i)<x)
        {
            y=*i;
            //x as B
            res-=B*x*(n-x-1);
            res-=A*y*(n-x-1);
            res-=C*up(x+1,n-1);
            //x as C
            res-=A*up(0,y-1);
            res-=B*y*y;
            res-=B*up(y+1,x-1);
            res-=A*y*(x-y-1>0?x-y-1:0);
            res-=C*x*(x-1);
        }
    //printf("%llu\n",res);
    //at least 2
    //int dig;
    //printf("%llu\n",res);
    for(x=0;x<n;x++) for(int i=1;i<=deg[x];i++)
    {
        //printf("%d %llu\n",x,sum[x][i]);
        //x has 2 edges out
        y=ed[x][i-1];
        if(y>x)
        {
            //x is A
            res+=A*x*(deg[x]-i); res+=B*y*(deg[x]-i);
            res+=C*(sum[x][deg[x]]-sum[x][i]);
            //printf("%d %d %llu\n",x,y,res);
        }
        else// if(y<x)
        {
            //x is B
            res+=B*x*(deg[x]-sz[x]); res+=A*y*(deg[x]-sz[x]);
            res+=C*(sum[x][deg[x]]-sum[x][sz[x]]);
            //x is C
            res+=C*x*(i-1); res+=B*y*(i-1);
            res+=A*sum[x][i-1];
        }
    }
    //printf("%llu\n",res);
    //at least 3
    /**for(int i=1;i<=n;i++)
    {
        if(deg[i]<=top)
        {
            for(int j=0;j<deg[i];j++)
            {
                for(int k=j+1;k<deg[i];k++)
                {
                    x=i; y=ed[x][j]; z=ed[x][k];
                    if(x>y)  swap(x,y);
                    if(x>z)  swap(x,z);
                    res-=A*x+B*y+C*z;
                }
            }
        }
        else    big.push_back(i);
    }*/
    for(x=0;x<n;x++) for(int i=0;i<deg[x];i++)
        if(deg[ed[x][i]]<deg[x] || (deg[ed[x][i]]==deg[x]&&ed[x][i]<x))
            add(x,ed[x][i]);
    memset(vis,-1,sizeof(vis));
    for(x=0;x<n;x++)
    {
        go(x);
        for(int i=in[x];i;i=e[i].lt)
        {
            y=e[i].to;
            for(int j=in[y];j;j=e[j].lt)
            {
                if(vis[e[j].to]==x)
                {
                    int xx=x,yy=y,zz=e[j].to;
                    if(xx>yy)    swap(xx,yy);
                    if(xx>zz)    swap(xx,zz);
                    if(yy>zz)    swap(yy,zz);
                    res-=A*xx+B*yy+C*zz;
                }
            }
        }
    }
    printf("%llu\n",res);
    //printf("%I64u\n",res);
    return 0;
}
/**
4 1
2 3 4
1 0
*/

 由于这个题调的我太过痛苦 写作业都写不进去

bug列举一下

1.我不会冒泡排序排三个数

2.(0,n-1)看成(1,n)重调系数非常蛋疼

3.初值就是忘了赋

4.系数和值少一个

5.手贱