二分图最大匹配题目汇总 POJ 1274、2239、3020、3715

 POJ 1274

#include <iostream>#include <string.h>#include <vector>#include <cstdio>using namespace std;vector<int> aa[205];int flag[205],visted[205];bool dfs(int x){for(int i=0;i<aa[x].size();++i){if(!visted[aa[x][i]]){  visted[aa[x][i]]=1;  if(!flag[aa[x][i]]||dfs(flag[aa[x][i]])){    flag[aa[x][i]]=x;return true;  }}}return false;}int main(){  int n,m;  while(~scanf("%d %d",&n,&m)){memset(aa,0,sizeof(aa));memset(flag,0,sizeof(flag));    int num,x;for(int i=1;i<=n;++i){  scanf("%d",&num);  while(num--){    scanf("%d",&x);aa[i].push_back(x);  }}int sum=0;for(int i=1;i<=n;++i){  memset(visted,0,sizeof(visted));  if(dfs(i))  sum++;}printf("%d\n",sum);  }  return 0;}

 POJ 2239

#include <iostream>#include <vector>#include <string.h>#include <cstdio>using namespace std;vector<int> ss[305];int flag[102],visit[102];bool dfs(int x){for(int i=0;i<ss[x].size();++i){if(!visit[ss[x][i]]){  visit[ss[x][i]]=1;  if(!flag[ss[x][i]]||dfs(flag[ss[x][i]])){//printf("x=%d  i=%d  flag=%d\n",x,i,flag[ss[x][i]]);    flag[ss[x][i]]=x;return true;  }}}return false;}int main(){  int n;  while(~scanf("%d",&n)){  int num,x,y;  memset(ss,0,sizeof(ss));  memset(flag,0,sizeof(flag));  for(int i=1;i<=n;++i){    scanf("%d",&num);while(num--){  scanf("%d %d",&x,&y);//day class  int xx=(x-1)*12+y;  ss[i].push_back(xx);}  }  int sum=0;  for(int i=1;i<=n;++i){    memset(visit,0,sizeof(visit));if(dfs(i))sum++;  }  printf("%d\n",sum);  }  return 0;}

 POJ 3020

法一：

#include<iostream>#include<cstdio>#include<vector>using namespace std;vector<int>map[400];//按奇偶性建二分图int t,n,m;int match[400],fy[400],ln,rn;int map1[41][11];//原始图int dirt[4][2]={0,1,0,-1,1,0,-1,0};int path(int u){for(int i=0,j;i<map[u].size();i++){j=map[u][i];if(!fy[j]){fy[j]=1;if(match[j]==-1||path(match[j])){match[j]=u;return 1;}}}return 0;}int main(){int i,j,x,y,ans;char tmp;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);ln=rn=0;for(i=1;i<=n;i++){getchar();for(j=1;j<=m;j++){scanf("%c",&tmp);if(tmp=='*'){if((i+j)&1)//奇点map1[i][j]=++ln;else//偶点map1[i][j]=++rn;}elsemap1[i][j]=0;}}for(i=1;i<=ln;i++)map[i].clear();for(i=1;i<=rn;i++)match[i]=-1;//建二分图for(i=1;i<=n;i++)for(j=1;j<=m;j++){if(!map1[i][j]||!((i+j)&1))continue;for(int k=0;k<4;k++){x=i+dirt[k][0];y=j+dirt[k][1];if(x<=0||x>n||y<=0||y>m||!map1[x][y])continue;map[map1[i][j]].push_back(map1[x][y]);}}//求最大匹配 最小点覆盖ans=0;for(i=1;i<=ln;i++){for(j=1;j<=rn;j++)fy[j]=0;if(path(i))ans++;}printf("%d\n",ln+rn-ans);}return 0;}

法二：

#include <iostream>#include <cstring>#include <cstdio>// 最大独立集//思路是出现*的坐标标为true，然后一每个为true的坐标为起点//用匈利亚算法确定是否有增广路存在。注意，//这样统计出来的数据是最大二分匹配的二倍，因为默认的看成了无向图。//result = 为true的结点数 - 最大二分匹配using namespace std;const int N = 50;bool g[N][N];bool usd[N][N];int lx[N][N], ly[N][N];int h, w;int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};bool can(int x, int y) {    int a, b, i;    for(i = 0; i < 4; ++i) {        a = x + dir[i][0];        b = y + dir[i][1];        if(!usd[a][b] && g[a][b]) {            usd[a][b] = true;            if((lx[a][b] == -1 && ly[a][b] == -1) || can(lx[a][b], ly[a][b])) {                lx[a][b] = x;                ly[a][b] = y;                return true;            }        }    }    return false;}int main() {    //freopen("data.in", "r", stdin);    int i, j, t;    int n, cnt;    char c;    scanf("%d", &t);    while(t--) {        scanf("%d%d", &h, &w);        getchar();        memset(g, false, sizeof(g));        for(n = 0, i = 1; i <= h; ++i) {            for(j = 1; j <= w; ++j) {                scanf("%c", &c);                //printf("%c", c);                if(c == '*')    {g[i][j] = true; n++;}               // printf("%d", g[i][j]);            }            getchar();           // cout << endl;        }        cnt = 0;        memset(lx, -1, sizeof(lx));        memset(ly, -1, sizeof(ly));        for(i = 1; i <= h; ++i) {            for(j = 1; j <= w; ++j) {                if(g[i][j]) {                    memset(usd, 0, sizeof(usd));                    if(can(i, j))   cnt++;                }            }        }        printf("%d\n", n - cnt/2);    }    return 0;}

法三：

#include <stdio.h>#include <stdlib.h>#include <string.h>int w,h;char scenario[45][15];int num[45][15];int map[410][410];bool isvisted[410];int match[410];int n;bool path(int u){for (int i = 1; i <= n; ++ i){if (map[u][i] && !isvisted[i]){isvisted[i] = true;if (match[i] == -1 || path(match[i])){match[i] = u;return true;}}}return false;}void buildMap(){memset(map, 0, sizeof(map));memset(num, 0, sizeof(num));n = 0;for (int i = 1; i <= h;  ++ i){for (int j = 1; j <= w; ++ j){if (scenario[i][j] == '*'){num[i][j] = ++n;}}}for (int i = 1; i <= h;  ++ i){for (int j = 1; j <= w; ++ j){if (num[i][j] > 0){if (num[i - 1][j] > 0){map[num[i][j]][num[i - 1][j]] = 1;}if (num[i + 1][j] > 0){map[num[i][j]][num[i + 1][j]] = 1;}if (num[i][j - 1] > 0){map[num[i][j]][num[i][j - 1]] = 1;}if (num[i][j + 1] > 0){map[num[i][j]][num[i][j + 1]] = 1;}//num[i][j] = 0;}}}}int main(){int t;scanf("%d", &t);while (t --){int count = 0;scanf("%d %d", &h, &w);memset(scenario, 0, sizeof(scenario));memset(match, -1, sizeof(match));for (int i = 1; i <= h; ++ i){scanf("%s", &scenario[i][1]);}buildMap();for (int i = 1; i <= n; ++ i ){memset(isvisted, false, sizeof(isvisted));if (path(i)){++count;}}printf("%d\n", n - count/2);}return 0;}

 POJ 3715

#include <iostream>//poj3715 博客里面有说明#include <algorithm>#include <cstdio>#include <cstring>using namespace std;#define maxn 250int n,m;int g[maxn][maxn];int a[maxn];int fugai[maxn],visit[maxn],prex[maxn],prey[maxn];void init(){memset(g,0,sizeof(g));memset(a,0,sizeof(a));memset(prex,-1,sizeof(prex));memset(prey,-1,sizeof(prey));memset(fugai,0,sizeof(fugai));}int canx(int x){if(fugai[x]==1)return 0;int i,j;for(i=0; i<n; i++)if(g[x][i] && visit[i]==0 && fugai[i]==0){visit[i]=1;if(prey[i]==-1 || canx(prey[i])){prex[x]=i;prey[i]=x;return 1;}}return 0;}int cany(int y){if(fugai[y]==1)return 0;int i,j;for(i=0; i<n; i++)if(g[y][i] && visit[i]==0 && fugai[i]==0){visit[i]=1;if(prex[i]==-1 || cany(prex[i])){prey[y]=i;prex[i]=y;return 1;}}return 0;}void solve(){int i,j,res=0;for(i=0; i<n; i++)if(a[i]==0)//匈牙利之用从二分图的一边，搜索{memset(visit,0,sizeof(visit));res+=canx(i);}printf("%d",res);int x,y;for(i=0; i<n; i++){if(a[i]==0)//覆盖集中的点可以是x的点也可以是y的点{if(prex[i]==-1)continue;//如果不是原来已经匹配的边y=prex[i];//从原来和他匹配的边找增广路prex[i]=-1;prey[y]=-1;fugai[i]=1;memset(visit,0,sizeof(visit));if(cany(y)==1)//如果找到增广路那么，这个点就不是覆盖集中的点{fugai[i]=0;//prex[i]=y;//prey[y]=i;}elseprintf(" %d",i);}else{if(prey[i]==-1)continue;x=prey[i];prey[i]=-1;prex[x]=-1;fugai[i]=-1;memset(visit,0,sizeof(visit));if(canx(x)==1){fugai[i]=0;//prey[i]=x;//prex[x]=i;}elseprintf(" %d",i);}}puts("");}int main(){    int t,i,j,k;scanf("%d", &t);while(t--){scanf("%d%d", &n,&m);init();int x,y;for(i=0; i<n; i++){scanf("%d", &x);a[i]=x;}for(i=0; i<m;i++){scanf("%d%d", &x,&y);if(a[x]!=a[y]){if(a[x]==1)swap(x,y);g[x][y]=1;//建图g[y][x]=1;}}solve();}}