Codeforces Round #360 (Div. 1)A - NP-Hard Problem

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链接：http://codeforces.com/contest/687/problem/A

题意：给定一个n个点的图，要求将所有的点分成2部分使得每一天边连着的两个点都不在同一部分中，无解输出-1。

分析：直接dfs01染色就是了。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=100010;const int MAX=1000000100;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=998244353;const int INF=1000000010;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;int tot,u[N],v[2*N],pre[2*N];void add(int x,int y) {    v[tot]=y;pre[tot]=u[x];u[x]=tot++;}int bo=1,q[N];void dfs(int x,int y) {    y==1 ? q[x]=2:q[x]=1;    for (int i=u[x];i!=-1;i=pre[i])    if (!q[v[i]]) dfs(v[i],q[x]);    else if (q[v[i]]==q[x]) bo=0;}int main(){    int i,n,m,x,y;    scanf("%d%d", &n, &m);    tot=0;memset(u,-1,sizeof(u));    for (i=1;i<=m;i++) {        scanf("%d%d", &x, &y);        add(x,y);add(y,x);    }    memset(q,0,sizeof(q));    for (i=1;i<=n;i++)    if (!q[i]) dfs(i,2);    if (!bo) printf("-1\n");    else {        m=0;        for (i=1;i<=n;i++)        if (q[i]==1) m++;        printf("%d\n", m);        for (i=1;i<=n;i++)        if (q[i]==1) printf("%d ", i);        printf("\n");        printf("%d\n", n-m);        for (i=1;i<=n;i++)        if (q[i]==2) printf("%d ", i);        printf("\n");    }    return 0;}

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