Codeforces 228E The Road to Berland is Paved With Good Intentions 枚举dfs判断可行性 || 并查集

题意：有n（1<=n<=100）个点的无向图，每条路的值是0或者1，King要把所有的边染色成1，但是修路的人很二逼，King让他去搞 i 点的时候

         他不仅把与 i 点相连的0变成1同时也把1变成0，问能不能找到一种方法使得所有0都变成1，spj。

题解：1 可以清楚的意识到一个点至多搞一次，枚举第一个点的状态然后同一个连通块的点的状态就可以定下来，dfs判定可行性即可。 但是图可能是不连通的唉。

         2 并查集，维护当前集合中根节点不取的时候其他节点与根节点的关系即可。

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dfs：

#include <iostream>#include <cstdio>#include <memory.h>#include <algorithm>using namespace std;const int maxn = 102;const int maxe = 10002;struct node{    int v,c;    int next;}edge[maxe];int head[maxn],ans[maxn],vis[maxn];bool record[maxn];int m,n,idx,top;void init(){    memset(head,-1,sizeof(head));    memset(record,false,sizeof(record));    idx = top = 0;    return;}void addedge(int u,int v,int c){    edge[idx].v = v;    edge[idx].c = c;    edge[idx].next = head[u];    head[u] = idx++;    edge[idx].v = u;    edge[idx].c = c;    edge[idx].next = head[v];    head[v] = idx++;    return;}void read(){    int u,v,w;    for(int i=0;i<m;i++)    {        scanf("%d %d %d",&u,&v,&w);        addedge(u,v,w);    }    return;}bool dfs(int st,int val){    vis[st] = val;    if(val == 1)    {        ans[top++] = st;    }    bool flag = true;    for(int i=head[st];i != -1;i=edge[i].next)    {        if(vis[edge[i].v] == -1)        {            if(dfs(edge[i].v , 1 ^ (edge[i].c ^ val)) == false)            {                flag = false;                break;            }        }        else        {            int tmp = (val + vis[edge[i].v]) % 2;            if(tmp == edge[i].c)            {                flag = false;                break;            }        }    }    return flag;}void out(){    printf("%d\n",top);    for(int i=0;i<top;i++)    {        if(i) printf(" ");        printf("%d",ans[i]);    }    puts("");    return;}void solve(){    bool flag = true;    top = 0;    for(int i=1;i<=n;i++)    {        int bj = top;        if(record[i] == false)        {            memset(vis,-1,sizeof(vis));            if(dfs(i , 0))            {                for(int j=1;j<=n;j++)                {                    if(vis[j] != -1)                    {                        record[j] = true;                    }                }            }            else            {                memset(vis,-1,sizeof(vis));                top = bj;                if(dfs(i , 1))                {                    for(int j=1;j<=n;j++)                    {                        if(vis[j] != -1)                        {                            record[j] = true;                        }                    }                }                else                {                    flag = false;                    break;                }            }        }    }    if(flag == false) puts("Impossible");    else out();    return;}int main(){    while(~scanf("%d %d",&n,&m))    {        init();        read();        solve();    }    return 0;}

并查集：

#include <iostream>#include <cstdio>#include <memory.h>using namespace std;const int maxn = 102;int father[maxn],dis[maxn],ans[maxn];int m,n;int find(int u){    if(u == father[u]) return father[u];    int t = father[u];    father[u] = find(father[u]);    dis[u] ^= dis[t];    return father[u];}void init(){    for(int i=1;i<=n;i++)    {        father[i] = i;        dis[i] = 0;    }    return;}void solve(){    int u,v,c;    bool flag = true;    while(m--)    {        scanf("%d %d %d",&u,&v,&c);        if(flag == false) continue;        int x = find(u);        int y = find(v);        if(x == y)        {            if((dis[u] ^ dis[v] ^ c) == 0)            {                flag = false;            }        }        else        {            father[y] = x;            dis[y] = 1 ^ dis[u] ^ dis[v] ^ c;        }    }    if(flag == false) puts("Impossible");    else    {        int top = 0;        for(int i=1;i<=n;i++)        {            find(i);            if(dis[i]) ans[top++] = i;        }        printf("%d\n",top);        for(int i=0;i<top;i++)        {            if(i) printf(" ");            printf("%d",ans[i]);        }        puts("");    }    return;}int main(){    while(~scanf("%d %d",&n,&m))    {        init();        solve();    }    return 0;}