HDU multi-School Joint Exercise Warm up 2 Bipartite Graph Matching

Warm up 2
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission (s): 656 Accepted Submission (s): 329

Problem Description
Some 1 × 2 dominoes are placed on a plane. each dominoe is placed either horizontally or vertically. it's guaranteed the dominoes in the same direction are not overlapped, but horizontal and vertical dominoes may overlap with each other. you task is to remove some dominoes, so that the remaining dominoes do not overlap with each other. now, tell me the maximum number of dominoes left on the board.
 

Input
There are multiple input cases.
The first line of each case are 2 integers: n (1 <= n <= 1000), m (1 <= m <= 1000 ), indicating the number of horizontal and vertical dominoes.
Then n lines follow, each line contains 2 integers x (0 <= x <= 100) and y (0 <= y <= 100), indicating the position of a horizontal dominoe. the dominoe occupies the grids of (x, y) and (x + 1, y ).
Then m lines follow, each line contains 2 integers x (0 <= x <= 100) and y (0 <= y <= 100), indicating the position of a horizontal dominoe. the dominoe occupies the grids of (x, y) and (x, y + 1 ).
Input ends with n = 0 and m = 0.
 

Output
For each test case, output the maximum number of remaining dominoes in a line.
 

Sample Input
2 3
0 0
0 3
0 1
1 1
1 3
4 5
0 1
0 2
3 1
2 2
0 0
1 0
2 0
4 1
3 2
0 0

Sample Output
4
6

Source
2013 Multi-University Training Contest 2
 

Recommend
Zhuyuanchen520
 

Question:
It turns out that I cannot create a graph... It is really not a solid foundation ..

It turns out to be a common largest independent set. The maximum independent set is defined as the selected point set. Any two edges are not connected. And because the question says that the question is not horizontally connected, the question is vertical and not vertical. This is a standard binary chart. Returns the largest independent set of X and Y.

When creating a graph, X takes the 0 to n-1 dominoes. Y takes the 0 to M-1 dominoes from the back... Speechless .. It turns out that I made a mistake in the 2-point graph ..

There are many solutions .. Search, greedy .. No more... Keep in mind the lessons of not creating a graph ..

I will provide the three template codes I have written ....

First 33 MS

#include <cstdio>#include <cstdlib>#include <cstring>#include <cstring>const int maxn=1111;int g[maxn][maxn];int cx[maxn],cy[maxn],vst[maxn];int nx,ny;int findpath(int u){    for(int v=0;v<ny;v++){        if(!vst[v]&&g[u][v]){            vst[v]=1;            if(cy[v]==-1||findpath(cy[v])){                cy[v]=u,cx[u]=v;                return 1;            }        }    }    return 0;}int MaxMatch(){    int ret=0;    memset(cx,-1,sizeof(cx));    memset(cy,-1,sizeof(cy));    for(int i=0;i<nx;i++)        if(cx[i]==-1){            memset(vst,0,sizeof(vst));            if(findpath(i))                ret++;        }    return ret;}struct node{    int x,y;}mx[maxn],my[maxn];int main(){    //freopen("1009.in","r",stdin);    int n,m;    while(scanf("%d %d",&n,&m)&&n+m){        for(int i=0,a,b;i<n;i++){            scanf("%d %d",&a,&b);            mx[i].x=a,mx[i].y=b;        }        for(int i=0,a,b;i<m;i++){            scanf("%d %d",&a,&b);            my[i].x=a,my[i].y=b;        }        memset(g,0,sizeof(g));        for(int i=0,x1,y1;i<n;i++){            x1=mx[i].x,y1=mx[i].y;            for(int j=0,x2,y2;j<m;j++){                x2=my[j].x,y2=my[j].y;                if(x1==x2&&y1==y2                 ||x1==x2&&y1==y2+1                 ||x1+1==x2&&y1==y2                 ||x1+1==x2&&y1==y2+1)                 g[i][j]=1;            }        }        nx=n,ny=m;        printf("%d\n",n+m-MaxMatch());    }    return 0;}

#include <cstdlib>#include <cstdio>#include <vector>#include <cstring>#include <iostream>#define clr(x,k) memset((x),(k),sizeof(x))#define foreach(it,c) for(vi::iterator it = (c).begin();it != (c).end();++it)using namespace std;typedef vector<int> vi;const int maxn=1000;vector<int> gx[maxn];int cx[maxn],cy[maxn],vst[maxn];int nx,ny;int findpath_Vector(int u){    foreach(it,gx[u]){        if(!vst[*it]){            vst[*it]=1;            if(cy[*it]==-1||findpath_Vector(cy[*it])){                    cx[u]=*it;                    cy[*it]=u;                    return 1;            }        }    }    return 0;}int maxMatch_Vector(){    int ret=0;    clr(cx,-1),clr(cy,-1);    for(int i=0;i<nx;i++)        if(cx[i]==-1){                clr(vst,0);                if(findpath_Vector(i))                    ret++;        }    return ret;}struct node{    int x,y;}mx[maxn],my[maxn];int main(){   // freopen("1009.in","r",stdin);    int n,m;    while(scanf("%d %d",&n,&m)&&n+m){        for(int i=0,a,b;i<n;i++){            scanf("%d %d",&a,&b);            mx[i].x=a,mx[i].y=b;        }        for(int i=0,a,b;i<m;i++){            scanf("%d %d",&a,&b);            my[i].x=a,my[i].y=b;        }        for(int i=0;i<n;i++)            gx[i].clear();        for(int i=0,x1,y1;i<n;i++){            x1=mx[i].x,y1=mx[i].y;            for(int j=0,x2,y2;j<m;j++){                x2=my[j].x,y2=my[j].y;                if(x1==x2&&y1==y2                 ||x1==x2&&y1==y2+1                 ||x1+1==x2&&y1==y2                 ||x1+1==x2&&y1==y2+1)                 gx[i].push_back(j);            }        }        nx=n,ny=m;        printf("%d\n",n+m-maxMatch_Vector());    }    return 0;}