Dichotomy Chart "template"

Binary diagram: The vertex of the original G can be classified into two sets X and Y, all of the two vertices of the border are exactly one of the set X, and the other belongs to the set Y, which is said to be a two-point figure.
Binary graph matching: given a binary graph G, in a sub-figure m of G, any two edges in M's edge set are not attached to the same vertex, that is, a vertex has at most one edge. is called M is a match.
Binary Graph Max match: The match with the most number of edges in the graph is called the maximum match of the graph.
The binary chart is perfectly matched: if all points are on the matching edge, then the maximum match is called the perfect match.
Two-figure multiple-match: The binary graph matches a one-to-two match, where an element in set Y is allowed to match multiple elements in the set X (typically with a maximum limit of n), but the elements in collection X can only match multiple elements in the set Y.
The best match of the binary graph: The binary graph is weighted, and the maximum matching of the total weights is found in the graph.
Minimum point coverage for a binary diagram: Minimum coverage requires a minimum number of points (either x or Y sets) to have each edge associated with at least one of the points.
Binary graph minimum point coverage = two min. graph maximum Match
Minimum side overlay of a binary diagram: Select the fewest edges to allow all points in the graph to be overwritten.
The minimum side overlay of the binary graph = the non-direction graph is repeated two times, ans = N-Maximum number of matches/2
Minimum path overlay for a DAG graph: Overlay a directed acyclic graph with as few disjoint simple paths as possible
(DAG) All vertices of G, which is the minimum path coverage problem for the DAG graph.
Minimum path overwrite = number of nodes n maximum number of matches
The maximum independent set of the binary graph: A number of points are selected in Figure g of N points, so that there is no edge between the several points 22, and the maximum value of the points is calculated.
Maximum number of independent sets of binary graphs = number of nodes (n)-Maximum number of matches (m)
Typically bidirectional, it is repeated two times, ans = N-Maximum number of matches/2

Binary graph maximum matching-Hungarian algorithm Dfs edition:

Const intMAXN = -;BOOLMAP[MAXN][MAXN];BOOLBMASK[MAXN];intNx,ny;intCX[MAXN],CY[MAXN];intFindpath (intu) { for(inti =1; I <= NY; ++i) {if(Map[u][i] &&!bmask[i]) {Bmask[i] =1;if(Cy[i] = =-1||                  Findpath (Cy[i])) {Cy[i] = u; Cx[u] = i;return 1; }          }      }return 0; }intMaxmatch () {intres =0; for(inti =1; I <= NX; ++i) Cx[i] =-1; for(inti =1; I <= NY; ++i) Cy[i] =-1; for(inti =1; I <= NX; ++i) {if(Cx[i] = =-1)          { for(intj =1; J <= NY; ++J) Bmask[j] =0;          Res + = Findpath (i); }      }returnRes//Returns the maximum number of matches}

Binary graph multiple matching:

Const intMAXN = -;//maximum number of verticesintMAP[MAXN][MAXN];//Two-minute chartBOOLMASK[MAXN];//array of flags when looking for an augmented pathintNx,ny,n;//nx Left collection vertex number, NY right collection vertex numberintVCY[MAXN];//vcy[i] Represents the number of vertices of the right collection I vertex matching to the left collectionintCY[MAXN][MAXN];//cy[i][j] Represents the first J element that matches the I vertex of the right collectionintLIMIT[MAXN];each vertex of the right collection matches the maximum number of left collection vertices///can also be limit, which represents the maximum number of common constraints that match the left collection vertex, and a binary search for the answer if the element is to be askedBOOLFindpath (intU//Search for augmented paths{ for(inti =1; I <=5; ++i) {if(Map[u][i] &&! Mask[i]) {Mask[i] =1;if(Vcy[i] < limit[i])//vcy[i] < limit{cy[i][vcy[i]++] = u;return true; } for(intj =0; J < Vcy[i]; ++J)//j < Vcy[i]{if(Findpath (Cy[i][j])) {Cy[i][j] = u;return true; }            }        }    }return false;}voidMulmatch ()//Find multiple matches{intAns =0;memset(VCY,0,sizeof(VCY)); for(inti =1; I <= N; ++i) {memset(Mask,0,sizeof(Mask)); Ans + = Findpath (i);//Calculate right to match the number of points        /* IF (!        Findpath (i)) return false; */}//return true;    if(Ans = = N)printf("T-shirts rock!\n");Else        printf("I ' d rather not wear a shirt anyway...\n");}

The best matching--km algorithm of binary graph:

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace STD;Const intMAXN = the;Const intINF =0xffffff0;intN,nx,ny;intLINK[MAXN],LX[MAXN],LY[MAXN],SLACK[MAXN];intVISX[MAXN],VISY[MAXN];//MarkintMAP[MAXN][MAXN];//Storage weight value//lx[],ly[] Top label; link[] Record matching valueintFindpath (intU//backtracking to find the optimal solution{Visx[u] =1; for(inti =1; I <= NY; ++i) {if(Visy[i])Continue;inttemp = Lx[u] + ly[i]-map[u][i];if(temp = =0)//if (map[u][i] = = Lx[u] + ly[i])//description is an equal sub-graph{Visy[i] =1;if(Link[i] = =-1|| Findpath (Link[i])) {Link[i] = u;return 1; }        }Else if(Slack[i] > Temp) slack[i] = temp; }return 0;}intKM ()//The best match for the highest weighted value{memset(Ly,0,sizeof(ly));memset(link,-1,sizeof(link)); for(inti =1; I <= NX; ++i) {Lx[i] =-inf;///for minimum right matching then lx[i] = INF         for(intj =1; J <= NY; ++J)if(Map[i][j] > Lx[i])//min Right match change symbolLx[i] = Map[i][j]; } for(inti =1; I <= NX; ++i) { for(intj =1; J <= NY; ++J) Slack[j] = INF; while(1)        {memset(VISX,0,sizeof(VISX));memset(Visy,0,sizeof(Visy));if(Findpath (i)) Break;intD = INF; for(intj =1; J <= NY; ++J)if(!visy[j] && d > slack[j]) d = slack[j];if(d = = INF)return; for(intj =1; J <= NX; ++J)if(Visx[j]) lx[j]-= D;//The minimum weight is changed to lx[j] + = D;             for(intj =1; J <= NY; ++J)if(Visy[j]) ly[j] + = D;//The minimum weight value is changed to ly[j]-= D;                ElseSLACK[J]-= D; }    }intres =0; for(inti =1; I <= NY; ++i)if(Link[i] >-1) Res + = Map[link[i]][i];returnRes//Output best match for maximum weights and}intMain () {intN while(~scanf("%d", &n)) {NX = NY = N; for(inti =1; I <= N; ++i) for(intj =1; J <= N; ++J)scanf("%d", &map[i][j]);printf("%d\n", KM ()); }return 0;}

Dichotomy Chart "template"