Hdu 4738 Caocao's Bridges (有重边无向图求桥)

2013杭州网络赛的第一题，坑点相当多，貌似这题导致15分钟没有队伍过题……

比赛时我负责这道题，WA9次才A。。。。。

从边数小于等于点数平方这点来看肯定有重边，需要处理。

图不连通，输出0.

如果取到的最小值是0的话，要输出1，表示要派一个人过去。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))

const int N=1005;
const int M=N*N*2;

struct Edge
{
	int v,next,val;
	bool sign;
}edge[M];

int head
,e;
int dfn
,low
;
int brg[M],cnt;   //存储桥的权值及数量
int top,dp;   //dp存储搜索深度
int n,m;

void Add (int u,int v,int val)
{
	edge[e].v = v ;
	edge[e].val = val ;
	edge[e].next = head[u] ;
	edge[e].sign = false ;
	head[u] = e ++ ;
}

void Init ()
{
	memset(head,-1,sizeof(head));
	e=cnt=0;
	memset(dfn,0,sizeof(dfn));
	memset(low,0,sizeof(low));
	memset(brg,0,sizeof(brg));
	top=dp=0;
}

void tarjan (int now ,int fa)
{
	dfn[now] = low[now] = ++dp;
	bool first =false ;
	for (int i = head[now] ; ~i ; i = edge[i].next)
	{
		int e = edge[i].v;
		if (edge[i].sign) continue;
		//无向图加双向边时必定相邻
		//处理后避免深搜的时候搜回去,经过重边再搜回来
		edge[i].sign = edge[i ^ 1].sign = true;  
		if (!dfn[e])
		{
			tarjan(e,now);
			low[now] = min(low[now] ,low[e]);
			if (dfn[now] < low[e])   //是桥
				brg[cnt++] = edge[i].val ;
		}
		else low[now] = min(low[now] , dfn[e]);
	}
}

int main ()
{
#ifdef ONLINE_JUDGE
#else
	freopen("read.txt","r",stdin);
#endif

    while (scanf("%d%d",&n,&m),n||m)
    {
		Init();
		int i,a,b,c;
		for (i=1;i<=m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			Add(a,b,c);
			Add(b,a,c);
		}
		int colornum=0;
        for (i=1;i<=n;i++)
        {
            top=dp=0;
            if (dfn[i]==0)
                tarjan (i,-1),colornum++;
        }
        if (colornum>1)//不连通
			printf("0\n");
		else if (cnt==0)  //没有桥
			printf("-1\n");
		else
		{
			sort(brg,brg+cnt);
			if (brg[0]==0)   //存在桥权值为0
				printf("1\n");
			else
				printf("%d\n",brg[0]);
		}
    }
    return 0;
}

2014-7-31更新一种写法

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#define min(x,y) ((x)<(y)?(x):(y))
#define max(x,y) ((x)>(y)?(x):(y))
using namespace std;

const int INF=0x3f3f3f3f;

const int nPoint=1005;
const int nEdges=1005*1005*2;
int brg[nEdges];

class BCC
{
public:
	struct Edge{
		int from, to,val, next;
		bool cut;  //是否为桥
	}edge[nEdges];

	int e,id,n;
	int head[nPoint],dfn[nPoint], low[nPoint];
	int bridgetop;
	int colornum, top; //双连通分量数,栈顶
	int color[nPoint],Stack[nPoint];
	bool iscut[nPoint];  //该点是否为割点
	int bri_cut;    //桥的数目

	void Add (int u, int v,int val){
		Edge E={u,v,val,head[u],false};
		edge[ e ] = E;
		head[u] = e++;
	}

	void Tarjan (int u, int pre)
	{
		dfn[u]=low[u]=++id;
		Stack[++top]=u;
		int child=0, flag=1;
		
		for (int i=head[u]; ~i; i=edge[i].next)
		{
			int v=edge[i].to;
			//if (v == pre) continue; //重边算一条的写法
			if (flag && v==pre)
			{//重边有效的写法
				flag = 0; 
				continue;
			}
			if (!dfn[v])
			{
				child++;
				Tarjan(v,u);
				low[u] = min(low[u], low[v]);
				if (low[v] >= dfn[u])
				{
					iscut[u] = true;  //是割点
					if (low[v]>dfn[u])
						edge[i].cut = edge[i^1].cut = true;  //是桥
				}
			}
			else low[u] = min(low[u], dfn[v]);
		}
		if (child == 1 && pre<0) //树根
			iscut[u] = false;
		if (low[u] == dfn[u])
		{
			colornum++;
			do
			{
				color[ Stack[top] ] = colornum;
			}while(Stack[top--] != u);
		}
	}

	void Init (int _n)
	{
		n=_n;
		memset(head, -1, sizeof(head));
		memset(dfn, 0, sizeof(dfn));
		memset(iscut, 0, sizeof(iscut));
		memset(color, -1, sizeof(color));
		bridgetop =e=id= 0;
		top = colornum = 0;
	}

	bool Deal ()
	{
		int i,flag=0;
		for (i=1; i<=n; i++) if (!dfn[i])
		{
			Tarjan(i, -1);
			flag++;
		}
		bri_cut = 0;
		for (i=0; i<e; i+=2)
			if (edge[i].cut)
			{
				brg[bri_cut]=edge[i].val;
				bri_cut++;
			}	
		if (flag==1) return true;
		return false;
	}
}ob;

int n,m;

int main ()
{
#ifdef ONLINE_JUDGE
#else
	freopen("read.txt","r",stdin);
#endif

    while (scanf("%d%d",&n,&m),n||m)
    {
		ob.Init(n);
		int i,a,b,c;
		for (i=1;i<=m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			ob.Add(a,b,c);
			ob.Add(b,a,c);
		}
		bool flag=ob.Deal();
		int cnt=ob.bri_cut;
		if (flag==false)//不连通
			printf("0\n");
		else if (cnt==0)  //没有桥
			printf("-1\n");
		else
		{
			sort(brg,brg+cnt);
			if (brg[0]==0)   //存在桥权值为0
				printf("1\n");
			else
				printf("%d\n",brg[0]);
		}
    }
    return 0;
}