題意：給你一個圖，問最多能添加多少條邊使圖仍為不是強連通圖，如果原圖是強連通輸出 ‘-1’分析：先把求出連通分量進行縮點，因為是求最多的添加邊，所以可以看成兩部分 x，y，只能一部分向另外一部分連邊，內部的就是完全圖，所以是x*（x+1）+x*y+y*（y+1）-M，只需要求出來出度或者入度為0的最少點的那個連通分量即可。**********************************************************************#include<stdio.h>
#include<string.h>
#include<algorithm>
using?namespace?std;

const?int?MAXN?=?1e5+5;
const?int?oo?=?1e9;

struct?Edge{int?v,?next;}e[MAXN];
int?Head[MAXN],?cnt;
void?AddEdge(int?u,?int?v)
{
????e[cnt].v?=?v;
????e[cnt].next?=?Head[u];
????Head[u]?=?cnt++;
}

int?dfn[MAXN],?low[MAXN],?Index;
int?Stack[MAXN],?top,?inStack[MAXN];
int?blg[MAXN],?bnt,?nblg[MAXN];///屬于哪個連通分量，連通分量里面有幾個點
int?outEdge[MAXN],?inEdge[MAXN];

void?InIt(int?N)
{
????cnt?=?Index?=?top?=?bnt?=?0;
????for(int?i=0;?i<=N;?i++)
????{
????????Head[i]?=?-1;
????????dfn[i]?=?0;
????????nblg[i]?=?0;
????????outEdge[i]?=?0;
????????inEdge[i]?=?0;
????}
}
void?Tarjan(int?u)
{
????int?v;

????low[u]?=?dfn[u]?=?++Index;
????Stack[++top]?=?u;
????inStack[u]?=?true;

????for(int?j=Head[u];?j!=-1;?j=e[j].next)
????{
????????v?=?e[j].v;
????????if(?!dfn[v]?)
????????{
????????????Tarjan(v);
????????????low[u]?=?min(low[u],?low[v]);
????????}
????????else?if(inStack[v]?==?true)
????????????low[u]?=?min(low[u],?dfn[v]);
????}

????if(low[u]?==?dfn[u])
????{
????????++bnt;
????????do
????????{
????????????v?=?Stack[top--];
????????????inStack[v]?=?false;
????????????blg[v]?=?bnt;
????????????nblg[bnt]++;
????????}
????????while(u?!=?v);
????}
}

int?main()
{
????int?T,?t=1;

????scanf("%d",?&T);

????while(T--)
????{
????????int?i,?j,?u,?v,?N,?M;

????????scanf("%d%d",?&N,?&M);

????????InIt(N);

????????for(i=0;?i<M;?i++)
????????{
????????????scanf("%d%d",?&u,?&v);
????????????AddEdge(u,?v);
????????}

????????for(i=1;?i<=N;?i++)
????????{
????????????if(?!dfn[i]?)
????????????????Tarjan(i);
????????}

????????for(i=1;?i<=N;?i++)
????????for(j=Head[i];?j!=-1;?j=e[j].next)
????????{
????????????v?=?e[j].v;
????????????if(blg[i]?!=?blg[v])
????????????{
????????????????inEdge[?blg[v]?]++;
????????????????outEdge[?blg[i]?]++;
????????????}
????????}

????????int?x,?y=oo;

????????for(i=1;?i<=bnt;?i++)
????????{
????????????if(!outEdge[i]?||?!inEdge[i])
????????????????y?=?min(y,?nblg[i]);
????????}

????????x?=?N-y;

????????if(bnt?==?1)
????????????printf("Case?%d:?-1\n",?t++);
????????else
????????????printf("Case?%d:?%lld\n",t++,?(long?long)x*(x-1)+x*y+y*(y-1)-M);
????}

????return?0;?

}

?

轉載于:https://www.cnblogs.com/liuxin13/p/4693700.html

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