Codeforce 505D - Mr. Kitayuta's Technology 弱聯通分量+拓撲排序

標籤：

對於有向圖M，若將其所有的邊轉化為無向邊，則得到其基圖M‘，若M’是聯通的，則稱有向圖M是弱聯通。

對於有向圖M，若圖中任意兩點u，v(u != v)均滿足u到v可達，v到u可達，則稱此圖為強聯通。

根據以上定義顯然可知，強聯通圖一定也滿足弱聯通。

此題首先我們需要找到其所有的弱聯通分量。

對於每一個弱聯通分量，設此弱聯通分量內點的個數為ans，如果此聯通分量無環，則需要的邊數為ans-1，若有環則為ans。

太挫了，這種題都不會了，怎麼變黃！！！

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <stack>#include <map>#pragma comment(linker, "/STACK:1024000000")#define EPS (1e-8)#define LL long long#define ULL unsigned long long#define _LL __int64#define INF 0x3f3f3f3f#define Mod 6000007//** I/O Accelerator Interface .. **/#define g (c=getchar())#define d isdigit(g)#define p x=x*10+c-'0'#define n x=x*10+'0'-c#define pp l/=10,p#define nn l/=10,ntemplate<class T> inline T& RD(T &x){    char c;    while(!d);    x=c-'0';    while(d)p;    return x;}template<class T> inline T& RDD(T &x){    char c;    while(g,c!='-'&&!isdigit(c));    if (c=='-')    {        x='0'-g;        while(d)n;    }    else    {        x=c-'0';        while(d)p;    }    return x;}inline double& RF(double &x)      //scanf("%lf", &x);{    char c;    while(g,c!='-'&&c!='.'&&!isdigit(c));    if(c=='-')if(g=='.')        {            x=0;            double l=1;            while(d)nn;            x*=l;        }        else        {            x='0'-c;            while(d)n;            if(c=='.')            {                double l=1;                while(d)nn;                x*=l;            }        }    else if(c=='.')    {        x=0;        double l=1;        while(d)pp;        x*=l;    }    else    {        x=c-'0';        while(d)p;        if(c=='.')        {            double l=1;            while(d)pp;            x*=l;        }    }    return x;}#undef nn#undef pp#undef n#undef p#undef d#undef gusing namespace std;vector<int>  vec[100010];vector<int> wvec[100010];int deg[100010];bool mark[100010];bool cir;int ans;struct Q{    int v,d;    bool operator < (const Q &a)const{        return a.d < d;    }};priority_queue<Q> q;void FindCir(){    Q f;    int i;    while(q.empty() == false)    {        f = q.top();        q.pop();        if(deg[f.v] < f.d)            continue;        if(f.d)        {            cir = true;            return ;        }        for(i = wvec[f.v].size()-1;i >= 0; --i)        {            --deg[wvec[f.v][i]];            q.push((Q){wvec[f.v][i],deg[wvec[f.v][i]]});        }    }}void FindAns(int site,int pre = -1){    if(mark[site])        return ;    mark[site] = true;    ans++;    q.push((Q){site,deg[site]});    for(int i = vec[site].size()-1;i >= 0; --i)    {        if(vec[site][i] != pre)            FindAns(vec[site][i],site);    }}int main(){    int u,v,i,n,m,sum;    scanf("%d %d",&n,&m);    memset(deg,0,sizeof(deg));    for(i = 1;i <= m; ++i)    {        scanf("%d %d",&u,&v);        wvec[u].push_back(v);        deg[v]++;        vec[u].push_back(v);        vec[v].push_back(u);    }    memset(mark,false,sizeof(mark));    sum = 0;    for(i = 1;i <= n; ++i)    {        if(mark[i] == false)        {            ans = 0;            while(q.empty() == false)                q.pop();            FindAns(i);            cir = false;            FindCir();            sum += cir ? ans : ans-1;        }    }    printf("%d\n",sum);    return 0;}

Codeforce 505D - Mr. Kitayuta's Technology 弱聯通分量+拓撲排序