一般图匹配带花树
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Appoint description:
Description
There is certain amount of night guards that are available to protect the local junkyard from possible junk robberies. These guards need to scheduled in pairs, so that each pair guards at different night. The junkyard CEO ordered you to write a program which given the guards characteristics determines the maximum amount of scheduled guards (the rest will be fired). Please note that each guard can be scheduled with only one of his colleagues and no guard can work alone.
Input
The first line of the input contains one number N ≤ 222 which is the amount of night guards. Unlimited number of lines consisting of unordered pairs ( i, j) follow, each such pair means that guard # i and guard # j can work together, because it is possible to find uniforms that suit both of them (The junkyard uses different parts of uniforms for different guards i.e. helmets, pants, jackets. It is impossible to put small helmet on a guard with a big head or big shoes on guard with small feet). The input ends with Eof.
Output
You should output one possible optimal assignment. On the first line of the output write the even number C, the amount of scheduled guards. Then output C/2 lines, each containing 2 integers ( i, j) that denote that i and j will work together.
Sample Input
3 1 2 2 3 1 3
output
2
1 2
题意为有若干组巡逻的保安可以搭配，输入第一行为搭配的数目，下面n行为可以匹配的关系，问最多可以形成多少对巡逻的保安，保证其中的任何一个保安只在其中的一个分组中，输出第一行为可以搭配的组数，下面若干
行为可以搭配的各组情况
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MAXN = 250; int N; //点的个数，点的编号从1到N bool Graph[MAXN][MAXN]; int Match[MAXN]; bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN]; int Head,Tail; int Queue[MAXN]; int Start,Finish; int NewBase; int Father[MAXN],Base[MAXN]; int Count;//匹配数，匹配对数是Count/2 void CreateGraph() { int u,v; memset(Graph,false,sizeof(Graph)); scanf("%d",&N); while(scanf("%d%d",&u,&v) == 2) { Graph[u][v] = Graph[v][u] = true; } } void Push(int u) { Queue[Tail] = u; Tail++; InQueue[u] = true; } int Pop() { int res = Queue[Head]; Head++; return res; } int FindCommonAncestor(int u,int v) { memset(InPath,false,sizeof(InPath)); while(true) { u = Base[u]; InPath[u] = true; if(u == Start) break; u = Father[Match[u]]; } while(true) { v = Base[v]; if(InPath[v])break; v = Father[Match[v]]; } return v; } void ResetTrace(int u) { int v; while(Base[u] != NewBase) { v = Match[u]; InBlossom[Base[u]] = InBlossom[Base[v]] = true; u = Father[v]; if(Base[u] != NewBase) Father[u] = v; } } void BloosomContract(int u,int v) { NewBase = FindCommonAncestor(u,v); memset(InBlossom,false,sizeof(InBlossom)); ResetTrace(u); ResetTrace(v); if(Base[u] != NewBase) Father[u] = v; if(Base[v] != NewBase) Father[v] = u; for(int tu = 1; tu <= N; tu++) if(InBlossom[Base[tu]]) { Base[tu] = NewBase; if(!InQueue[tu]) Push(tu); } } void FindAugmentingPath() { memset(InQueue,false,sizeof(InQueue)); memset(Father,0,sizeof(Father)); for(int i = 1; i <= N; i++) Base[i] = i; Head = Tail = 1; Push(Start); Finish = 0; while(Head < Tail) { int u = Pop(); for(int v = 1; v <= N; v++) if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v)) { if((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0)) BloosomContract(u,v); else if(Father[v] == 0) { Father[v] = u; if(Match[v] > 0) Push(Match[v]); else { Finish = v; return; } } } } } void AugmentPath() { int u,v,w; u = Finish; while(u > 0) { v = Father[u]; w = Match[v]; Match[v] = u; Match[u] = v; u = w; } } void Edmonds() { memset(Match,0,sizeof(Match)); for(int u = 1; u <= N; u++) if(Match[u] == 0) { Start = u; FindAugmentingPath(); if(Finish > 0)AugmentPath(); } } void PrintMatch() { Count = 0; for(int u = 1; u <= N; u++) if(Match[u] > 0) Count++; printf("%d\n",Count); for(int u = 1; u <= N; u++) if(u < Match[u]) printf("%d %d\n",u,Match[u]); } int main() { CreateGraph();//建图 Edmonds();//进行匹配 PrintMatch();//输出匹配数和匹配 return 0; }
Appoint description:
Description
A new season of Touhou M-1 Grand Prix is approaching. Girls in Gensokyo cannot wait for participating it. Before the registration, they have to decide which combination they are going to compete as. Every girl in Gensokyo is both a boke (funny girl) and a tsukkomi (straight girl). Every candidate combination is made up of two girls, a boke and a tsukkomi. A girl may belong to zero or more candidate combinations, but one can only register as a member of one formal combination. The host of Touhou M-1 Grand Prix hopes that as many formal combinations as possible can participate in this year. Under these constraints, some candidate combinations are actually redundant as it\‘s impossible to register it as a formal one as long as the number of formal combinations has to be maximized. So they want to figure out these redundant combinations and stop considering about them.
Input
There are multiple test cases. Process to the End of File.
The first line of each test case contains two integers: 1 ≤ N ≤ 40 and 1 ≤ M ≤ 123, where N is the number of girls in Gensokyo, and M is the number of candidate combinations. The following M lines are M candidate combinations, one by each line. Each combination is represented by two integers, the index of the boke girl 1 ≤ B i ≤ N and the index of the tsukkomi girl 1 ≤ T i ≤ N, where B i != T i.
The first line of each test case contains two integers: 1 ≤ N ≤ 40 and 1 ≤ M ≤ 123, where N is the number of girls in Gensokyo, and M is the number of candidate combinations. The following M lines are M candidate combinations, one by each line. Each combination is represented by two integers, the index of the boke girl 1 ≤ B i ≤ N and the index of the tsukkomi girl 1 ≤ T i ≤ N, where B i != T i.
Output
For each test case, output the number of redundant combinations in the
first line. Then output the space-separated indexes of the redundant
combinations in ascending order in the second line.
Sample Input
4 4 1 3 2 3 2 4 3 1 6 6 1 2 3 2 3 4 5 2 5 4 5 6
Sample Output
1
2
3
2 4 5
题意为输入第一行为两个整数nm，n代表有n个打算参加跳舞的女孩，m代表可一匹配的队伍，
我们想要最大化可以组合的队伍，输出最大化的组合之后，其他没有必要的组合，输出他们的序号
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MAXN = 50; int N; //点的个数，点的编号从1到N bool Graph[MAXN][MAXN]; int Match[MAXN]; bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN]; int Head,Tail; int Queue[MAXN]; int Start,Finish; int NewBase; int Father[MAXN],Base[MAXN]; int Count; void Push(int u) { Queue[Tail] = u; Tail++; InQueue[u] = true; } int Pop() { int res = Queue[Head]; Head++; return res; } int FindCommonAncestor(int u,int v) { memset(InPath,false,sizeof(InPath)); while(true) { u = Base[u]; InPath[u] = true; if(u == Start) break; u = Father[Match[u]]; } while(true) { v = Base[v]; if(InPath[v])break; v = Father[Match[v]]; } return v; } void ResetTrace(int u) { int v; while(Base[u] != NewBase) { v = Match[u]; InBlossom[Base[u]] = InBlossom[Base[v]] = true; u = Father[v]; if(Base[u] != NewBase) Father[u] = v; } } void BloosomContract(int u,int v) { NewBase = FindCommonAncestor(u,v); memset(InBlossom,false,sizeof(InBlossom)); ResetTrace(u); ResetTrace(v); if(Base[u] != NewBase) Father[u] = v; if(Base[v] != NewBase) Father[v] = u; for(int tu = 1; tu <= N; tu++) if(InBlossom[Base[tu]]) { Base[tu] = NewBase; if(!InQueue[tu]) Push(tu); } } void FindAugmentingPath() { memset(InQueue,false,sizeof(InQueue)); memset(Father,0,sizeof(Father)); for(int i = 1;i <= N;i++) Base[i] = i; Head = Tail = 1; Push(Start); Finish = 0; while(Head < Tail) { int u = Pop(); for(int v = 1; v <= N; v++) if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v)) { if((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0)) BloosomContract(u,v); else if(Father[v] == 0) { Father[v] = u; if(Match[v] > 0) Push(Match[v]); else { Finish = v; return; } } } } } void AugmentPath() { int u,v,w; u = Finish; while(u > 0) { v = Father[u]; w = Match[v]; Match[v] = u; Match[u] = v; u = w; } } void Edmonds() { memset(Match,0,sizeof(Match)); for(int u = 1; u <= N; u++) if(Match[u] == 0) { Start = u; FindAugmentingPath(); if(Finish > 0)AugmentPath(); } } int getMatch() { Edmonds(); Count = 0; for(int u = 1; u <= N;u++) if(Match[u] > 0) Count++; return Count/2; } bool g[MAXN][MAXN]; pair<int,int>p[150]; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int m; while(scanf("%d%d",&N,&m)==2) { memset(g,false,sizeof(g)); memset(Graph,false,sizeof(Graph)); int u,v; for(int i = 1;i <= m;i++) { scanf("%d%d",&u,&v); p[i] = make_pair(u,v); g[u][v] = true; g[v][u] = true; Graph[u][v] = true; Graph[v][u] = true; } int cnt0 = getMatch(); //cout<<cnt0<<endl; vector<int>ans; for(int i = 1;i <= m;i++) { u = p[i].first; v = p[i].second; memcpy(Graph,g,sizeof(g)); for(int j = 1;j <= N;j++) Graph[j][u] = Graph[u][j] = Graph[j][v] = Graph[v][j] = false; int cnt = getMatch(); //cout<<cnt<<endl; if(cnt < cnt0-1) ans.push_back(i); } int sz = ans.size(); printf("%d\n",sz); for(int i = 0;i < sz;i++) { printf("%d",ans[i]); if(i < sz-1)printf(" "); } printf("\n"); } return 0; }
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