1099. Work Scheduling
Time limit: 0.5 second
Memory limit: 64 MB
Memory limit: 64 MB
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 | output |
---|---|
3 1 2 2 3 1 3 |
2 1 2
|
模板题!!!
以下是模板:
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013/8/21 22:56:05 4 File Name :F:\2013ACM练习\专题学习\图论\一般图匹配带花树\URAL1099.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 21 const int MAXN = 250; 22 int N; //点的个数,点的编号从1到N 23 bool Graph[MAXN][MAXN]; 24 int Match[MAXN]; 25 bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN]; 26 int Head,Tail; 27 int Queue[MAXN]; 28 int Start,Finish; 29 int NewBase; 30 int Father[MAXN],Base[MAXN]; 31 int Count;//匹配数,匹配对数是Count/2 32 void CreateGraph() 33 { 34 int u,v; 35 memset(Graph,false,sizeof(Graph)); 36 scanf("%d",&N); 37 while(scanf("%d%d",&u,&v) == 2) 38 { 39 Graph[u][v] = Graph[v][u] = true; 40 } 41 } 42 void Push(int u) 43 { 44 Queue[Tail] = u; 45 Tail++; 46 InQueue[u] = true; 47 } 48 int Pop() 49 { 50 int res = Queue[Head]; 51 Head++; 52 return res; 53 } 54 int FindCommonAncestor(int u,int v) 55 { 56 memset(InPath,false,sizeof(InPath)); 57 while(true) 58 { 59 u = Base[u]; 60 InPath[u] = true; 61 if(u == Start) break; 62 u = Father[Match[u]]; 63 } 64 while(true) 65 { 66 v = Base[v]; 67 if(InPath[v])break; 68 v = Father[Match[v]]; 69 } 70 return v; 71 } 72 void ResetTrace(int u) 73 { 74 int v; 75 while(Base[u] != NewBase) 76 { 77 v = Match[u]; 78 InBlossom[Base[u]] = InBlossom[Base[v]] = true; 79 u = Father[v]; 80 if(Base[u] != NewBase) Father[u] = v; 81 } 82 } 83 void BloosomContract(int u,int v) 84 { 85 NewBase = FindCommonAncestor(u,v); 86 memset(InBlossom,false,sizeof(InBlossom)); 87 ResetTrace(u); 88 ResetTrace(v); 89 if(Base[u] != NewBase) Father[u] = v; 90 if(Base[v] != NewBase) Father[v] = u; 91 for(int tu = 1; tu <= N; tu++) 92 if(InBlossom[Base[tu]]) 93 { 94 Base[tu] = NewBase; 95 if(!InQueue[tu]) Push(tu); 96 } 97 } 98 void FindAugmentingPath() 99 { 100 memset(InQueue,false,sizeof(InQueue)); 101 memset(Father,0,sizeof(Father)); 102 for(int i = 1;i <= N;i++) 103 Base[i] = i; 104 Head = Tail = 1; 105 Push(Start); 106 Finish = 0; 107 while(Head < Tail) 108 { 109 int u = Pop(); 110 for(int v = 1; v <= N; v++) 111 if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v)) 112 { 113 if((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0)) 114 BloosomContract(u,v); 115 else if(Father[v] == 0) 116 { 117 Father[v] = u; 118 if(Match[v] > 0) 119 Push(Match[v]); 120 else 121 { 122 Finish = v; 123 return; 124 } 125 } 126 } 127 } 128 } 129 void AugmentPath() 130 { 131 int u,v,w; 132 u = Finish; 133 while(u > 0) 134 { 135 v = Father[u]; 136 w = Match[v]; 137 Match[v] = u; 138 Match[u] = v; 139 u = w; 140 } 141 } 142 void Edmonds() 143 { 144 memset(Match,0,sizeof(Match)); 145 for(int u = 1; u <= N; u++) 146 if(Match[u] == 0) 147 { 148 Start = u; 149 FindAugmentingPath(); 150 if(Finish > 0)AugmentPath(); 151 } 152 } 153 void PrintMatch() 154 { 155 Count = 0; 156 for(int u = 1; u <= N;u++) 157 if(Match[u] > 0) 158 Count++; 159 printf("%d\n",Count); 160 for(int u = 1; u <= N; u++) 161 if(u < Match[u]) 162 printf("%d %d\n",u,Match[u]); 163 } 164 int main() 165 { 166 CreateGraph();//建图 167 Edmonds();//进行匹配 168 PrintMatch();//输出匹配数和匹配 169 return 0; 170 }
来源:https://www.cnblogs.com/kuangbin/p/3278621.html