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 Endusing 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;elseaddedge(i,end, ch[i]>>1);}if(sum != dinic(start, end))printf("Poor boy!\n");else printf("Well done!\n");}}