二分图判断

来源：互联网发布：易语言网络爬虫编辑：程序博客网时间：2024/06/05 12:48

https://gcj-prod.appspot.com/codejam/contest/2933486/dashboard

bfs看是否有染色矛盾，两个都没染过色说明这是另一个图，再次bfs

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <string>#include <map>using namespace std;int ca,n;const int maxn = 11000;map <string,int> mp;string a[maxn],b[maxn];bool vid[maxn*2];int tot,head[maxn*2],col[maxn*2];struct node{    int v,next;}edges[maxn*2];void add(int u,int v){    edges[tot].v = v;edges[tot].next = head[u];head[u] = tot++;    edges[tot].v = u;edges[tot].next = head[v];head[v] = tot++;}void init(){    scanf("%d",&n);    for(int i = 0 ; i <= 2*n+1 ; i++)  head[i] = -1,col[i] = 0;    mp.clear();    for(int i =  1 ; i <= n ; i++){        cin >> a[i]>>b[i];        if(mp[a[i]] == 0)            mp[a[i]] = i*2;        if(mp[b[i]] == 0)            mp[b[i]] = i*2+1;        add(mp[a[i]],mp[b[i]]);    }}int bfs(int ka,int kb){    queue<int>q;    q.push(ka);    col[ka] = 1;    while(!q.empty()){        int tmp = q.front();q.pop();       // cout <<"asd "<<endl;        for(int k = head[tmp] ; k != -1; k = edges[k].next){            int v = edges[k].v;            //cout<<"v =  "<<v << "  col[v] = "<<col[v]<<" tmp = "<<tmp <<"  col[tmp = "<<col[tmp]<<endl;            if(col[v]!=0 && col[v] == col[tmp])                return -1;            else if(col[v] == 0){                q.push(v);               // cout << "V = "<<v<<endl;                col[v] = col[tmp] == 1? 2:1;            }        }    }    return 1;}void sov(){    for(int i = 1; i <= n ; i++){        int ka = mp[a[i]],kb = mp[b[i]];        if(col[ka] == 0 && col[kb] == 0){            if(bfs(ka,kb) == -1){                printf("Case #%d: No\n",ca);                return;            }        }    }    printf("Case #%d: Yes\n",ca);}int main(){    int T;    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    scanf("%d",&T);    for(ca = 1; ca <= T ;ca++){        init();        sov();    }}

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