Hdu2236 no question II Max matching + Binary Search

You can understand the meaning of the question in Chinese.

When we see "different rows and columns", we think we should use binary matching. The difference between the maximum value and the minimum value is minimized by the lower limit and upper limit of enumeration edge.

The enumeration process is like this. In the input process, you can record the maximum and minimum weights Min. The maximum value of their edge weight difference is right = max-min, and the minimum value is left = 0. Then, we constantly increase the lower limit of the edge and find the difference value of the edge weight. If we can get a perfect match (the matching number is equal to N), we will record the difference. Final output. This search process is similar to the largest stream + binary search.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<algorithm>  5 #include<queue>  6 using namespace std;  7 const int N=105,INF=0x3f3f3f3f;  8 int Map[N][N],cx[N],cy[N],dx[N],dy[N];  9 bool bmask[N],bmap[N][N]; 10 int nx,ny,dis,ans; 11 bool searchpath() 12 { 13     queue<int> q; 14     dis=INF; 15     memset(dx,-1,sizeof(dx)); 16     memset(dy,-1,sizeof(dy)); 17     for(int i=1;i<=nx;i++) 18     { 19         if(cx[i]==-1){ q.push(i); dx[i]=0; } 20         while(!q.empty()) 21         { 22             int u=q.front(); q.pop(); 23             if(dx[u]>dis) break; 24             for(int v=1;v<=ny;v++) 25             { 26                 if(bmap[u][v]&&dy[v]==-1) 27                 { 28                     dy[v]= dx[u] + 1; 29                     if(cy[v]==-1) dis=dy[v]; 30                     else 31                     { 32                         dx[cy[v]]= dy[v]+1; 33                         q.push(cy[v]); 34                     } 35                 } 36             } 37         } 38     } 39     return dis!=INF; 40 } 41 int findpath(int u) 42 { 43     for(int v=1;v<=ny;v++) 44     { 45         if(!bmask[v]&&bmap[u][v]&&dy[v]==dx[u]+1) 46         { 47             bmask[v]=1; 48             if(cy[v]!=-1&&dy[v]==dis) continue; 49             if(cy[v]==-1||findpath(cy[v])) 50             { 51                 cy[v]=u; cx[u]=v; 52                 return 1; 53             } 54         } 55     } 56     return 0; 57 } 58 void maxmatch() 59 { 60     ans=0; 61     memset(cx,-1,sizeof(cx)); 62     memset(cy,-1,sizeof(cy)); 63     while(searchpath()) 64     { 65         memset(bmask,0,sizeof(bmask)); 66         for(int i=1;i<=nx;i++) 67             if(cx[i]==-1) ans+=findpath(i); 68     } 69 } 70 void init() 71 { 72     memset(bmap,0,sizeof(bmap)); 73 } 74  75 int main() 76 { 77     //freopen("test.txt","r",stdin); 78     int i,j,k,n,cas,a,b; 79     scanf("%d",&cas); 80     while(cas--) 81     { 82         scanf("%d",&n); 83         a=100,b=0; 84         for(i=1;i<=n;i++) 85             for(j=1;j<=n;j++){ 86                 scanf("%d",&Map[i][j]); 87                 a=min(a,Map[i][j]); 88                 b=max(b,Map[i][j]); 89             } 90         nx=ny=n; 91         int L=0,R=b-a,mid,flag,res; 92         while(L<=R) 93         { 94             mid=(L+R)/2; 95             flag=0; 96             for(k=a;k<=b;k++) 97             { 98                 for(i=1;i<=n;i++) 99                     for(j=1;j<=n;j++)100                         if(Map[i][j]>=k&&Map[i][j]<=k+mid) bmap[i][j]=1;101                         else bmap[i][j]=0;102                 maxmatch();103                 if(ans==n) {flag=1;break;}104             }105             if(flag) {res=mid;R=mid-1;}106             else  L=mid+1;107         }108         printf("%d\n",res);109     }110     return 0;111 }

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Hdu2236 no question II Max matching + Binary Search