HDU 3472 混合图欧拉回路 + 网络流

九野的博客，转载请注明出处：http://blog.csdn.net/acmmmm/article/details/13799337

题意：

T个测试数据

n串字符 能否倒过来用(1表示能倒着用)

问能否把所有字符串 首尾相接

欧拉回路是图G中的一个回路，经过每条边有且仅一次，称该回路为欧拉回路。具有欧拉回路的图称为欧拉图，简称E图。

混合图就是边集中有有向边和无向边同时存在。这时候需要用网络流建模求解。

不能倒着用就是有向边，能倒着用就是无向边

http://yzmduncan.iteye.com/blog/1149049

.

欧拉回路要求出度=入度 ，因此若出度与入度 差为奇数，一定没有欧拉回路 ch[i]%1 必须==0

用并查集判断每个字母的祖先是否相同(（不相同则图不连通）

对于所有的点，总入度 + 总出度 = 0 【1】

ch[i]表示字母i+‘a' 的入度-出度的值

由【1】式推出 所有 ch[i] < 0的点 + ch[i]>0 的点 = 0

所以 让所有ch[i]<0 点与源点建边 ，权值为 -ch[i]

ch[i]>0 与汇点建边，权值为 ch[i]

若dinic满流，表示上述式子成立，则存在回路

#include <stdio.h>
#include <string.h>
#include <queue>
#define inf 10000
#define ll short
#define end End

using namespace std;

struct Edge{
	short from, to, cap, nex;
}edge[1007];
short head[28], edgenum;
void addedge(short u, short v, short cap){
	Edge E ={u, v, cap, head[u]};
	edge[edgenum] = E;
	head[u] = edgenum++;
	Edge E_ = {v,u,0,head[v]};
	edge[edgenum] = E_;
	head[v] = edgenum++;
}

short sign[28];
bool BFS(short from, short to){
	memset(sign, -1, sizeof(sign));
	sign[from] = 0;

	queue<short>q;
	q.push(from);
	while( !q.empty() ){
		int u = q.front(); q.pop();
		for(short i = head[u]; i!=-1; i = edge[i].nex)
		{
			short v = edge[i].to;
			if(sign[v]==-1 && edge[i].cap)
			{
				sign[v] = sign[u] + 1, q.push(v);
				if(sign[to] != -1)return true;
			}
		}
	}
	return false;
}
short Stack[4], top, cur[4];
short dinic(short from, short to){
	short ans = 0;
	while( BFS(from, to) )
	{
		memcpy(cur, head, sizeof(head));
		short u = from;		top = 0;
		while(1)
		{
			if(u == to)
			{
				short flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
				for(short i = 0; i < top; i++)
					if(flow > edge[ Stack[i] ].cap)
					{
						flow = edge[Stack[i]].cap;
						loc = i;
					}

					for(short i = 0; i < top; i++)
					{
						edge[ Stack[i] ].cap -= flow;
						edge[Stack[i]^1].cap += flow;
					}
					ans += flow;
					top = loc;
					u = edge[Stack[top]].from;
			}
			for(short i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
				if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
			if(cur[u] != -1)
			{
				Stack[top++] = cur[u];
				u = edge[ cur[u] ].to;
			}
			else
			{
				if( top == 0 )break;
				sign[u] = -1;
				u = edge[ Stack[--top] ].from;
			}
		}
	}
	return ans;
}
ll n;
char ss[22];
short ch[27], f[27];
bool use[27];
short start, end;

short find(short x){return x==f[x]?x:(f[x]=find(f[x]));}
void Union(short x, short y){
	short fx = find(x), fy = find(y);
	short temp = fx; if(fx>fy){fx=fy;fy=temp;}
	f[fx] = fy;
}
int main(){
	short T, i, j, Cas = 1; scanf("%d",&T);
	while(T--)
	{
		memset(ch, 0, sizeof(ch));
		memset(use,0, sizeof(use));
		for(i=0;i<27;i++)f[i] = i;
		memset(head, -1,sizeof(head)), edgenum = 0;

		scanf("%d",&n);
		for(i = 0; i < n; i++)
		{
			scanf("%s %d",ss,&j);
			int a = ss[0]-'a', b = ss[strlen(ss)-1] - 'a';
			ch[a]++, ch[b]--;
			use[a] = use[b] = true;
			Union(a,b);

			if(j)addedge(a,b,1);
		}
		printf("Case %d: ",Cas++);
		bool ok = true;
		for(i=0;i<26;i++)
			if(use[i])
			{
				j = i;
				for(i++;i<26;i++)
					if(use[i] && find(j)!=find(i))ok = false;
				break;
			}
		short num = 0;
		for(i=0;i<26;i++)if(use[i] &&  ch[i]%2)
		{
			num++;
			if(ch[i]<0)start = i;
			else end = i;
		}
		if(num == 1 || num>2)ok = false;
		if(!ok){ printf("Poor boy!\n"); continue;}
		if(num == 2)addedge(end, start, 1);
		start = 26, end = 27;
		short sum = 0;
		for(i=0;i<26;i++)if(ch[i] && use[i] && i!=end && i!=start)
		{
			if(ch[i]<0)
				addedge(start,i,-ch[i]/2), sum-=ch[i]/2;
			else
				addedge(i,end, ch[i]>>1);
		}
		if(sum != dinic(start, end))
			printf("Poor boy!\n");
		else
			printf("Well done!\n");

	}
}