The Accomodation of Students

　　　　　　　　　　　　　　Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　Total Submission(s): 7086????Accepted Submission(s): 3167

Problem Description There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

Calculate the maximum number of pairs that can be arranged into these double rooms.

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Input For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

Proceed to the end of file.

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Output If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.

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Sample Input 4 4 1 2 1 3 1 4 2 3 6 5 1 2 1 3 1 4 2 5 3 6

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Sample Output No 3

題目：一些學生之間是朋友關系（關系不能傳遞），現在要將一堆學生分成兩堆,使得在同一堆的學生之間沒有朋友關系。如果不可以輸出“No”，可以的話輸出最多可以分出幾對小盆友。

思路：

我們先二分圖染色，若能被染成兩部分的話說明可以被分成兩部分，然后再在我們分出的圖上面跑最大匹配。若不能被染成兩部分直接輸出no

代碼：

#include<queue> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 510 using namespace std; bool flag,vis[N]; int n,m,x,y,tot,ans,col[N],girl[N],head[N],map[N][N]; queue<int>q; struct Edge {int from,to,next; }edge[N*N]; int read() {int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f; } int add(int x,int y) {tot++;edge[tot].to=y;edge[tot].next=head[x];head[x]=tot; } int find(int x) {for(int i=1;i<=n;i++){if(!vis[i]&&map[x][i]){vis[i]=true;if(girl[i]==-1||find(girl[i])){girl[i]=x; return 1;}}}return 0; } int color(int s) {queue<int>q;q.push(s); col[s]=0;while(!q.empty()){int x=q.front();for(int i=head[x];i;i=edge[i].next){int t=edge[i].to;if(col[t]!=-1){if(col[t]==col[x]) {flag=true; return 1;}}else{col[t]=col[x]^1;q.push(t);}}q.pop();}return 0; } int main() {while(scanf("%d%d",&n,&m)!=EOF){ans=0;flag=false;tot=0;memset(map,0,sizeof(map));memset(col,-1,sizeof(col));memset(edge,0,sizeof(edge));memset(head,0,sizeof(head));for(int i=1;i<=m;i++){x=read(),y=read();map[x][y]=1;add(x,y),add(y,x);}for(int i=1;i<=n;i++)if(col[i]==-1){if(color(i)) break;} if(flag) {printf("No\n"); continue;}memset(girl,-1,sizeof(girl));for(int i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(find(i)) ans++;}printf("%d\n",ans);}return 0; }

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轉載于:https://www.cnblogs.com/z360/p/7435411.html

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