【ZJOI2017 Round1练习】D2T1 river（二分图）

题意：

思路：这道题并没有官方题解

没有羊驼在所有三元组中出现就是NO

现在考虑不少于1只的情况

删去其中一只，我们得到了两组点和一些边

我们只要判断这是否为一张二分图，使用暴力染色的方法就有60分了

#include<cstdio>
#include<algorithm>
#define N 10010
#define M 50010
using namespace std;
int T,n,m,edgenum,u,v,ok,isok,root;
int f[N],vet[M],next[M],head[N],a[M],b[M],c[M],vis[N],flag[N],col[N];
void add(int u,int v)
{
    vet[++edgenum]=v;
    next[edgenum]=head[u];
    head[u]=edgenum;
}
void dfs(int u)
{
    vis[u]=1;
    for (int e=head[u];e;e=next[e])
    {
        int v=vet[e];
        if (flag[v]) continue;
        if (vis[v])
        {
            if (col[v]==col[u]) ok=0;
        }else
        {
            col[v]=col[u]^1;
            dfs(v);
        }
    }
}
int main()
{
    freopen("river.in","r",stdin);
    freopen("river.out","w",stdout);
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        for (int i=1;i<=n;i++) f[i]=0;
        for (int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&a[i],&b[i],&c[i]);
            f[a[i]]++;f[b[i]]++;f[c[i]]++;
        }
        root=-1;
        for (int i=1;i<=n;i++) if (f[i]==m) root=i;
        if (root==-1){puts("no");continue;}
        edgenum=0;
        for (int i=1;i<=n;i++) head[i]=0;
        for (int i=1;i<=m;i++)
        {
            if (a[i]==root){u=b[i];v=c[i];}
            if (b[i]==root){u=a[i];v=c[i];}
            if (c[i]==root){u=a[i];v=b[i];}
            //ed[i].x=u;ed[i].y=v;
            add(u,v);add(v,u);
        }
        isok=0;
        //printf("%d
",root);
        for (int i=1;i<=n;i++) if (i!=root)
        {
            if (isok) break;
            for (int j=i+1;j<=n;j++) if (j!=root)
            {
                //printf("%d %d
",i,j);
                for (int k=1;k<=n;k++) vis[k]=col[k]=0;
                flag[i]=flag[j]=1;
                ok=1;
                for (int k=1;k<=n;k++) if (!vis[k]&&k!=i&&k!=j&&k!=root)
                    dfs(k);
                if (ok) isok=1;
                flag[i]=flag[j]=0;
                if (isok) break;
            }
        }
        if (isok) puts("yes");else puts("no");
    }
}

至于标程……谁看得懂呢……貌似是暴力加了点优化……

#include<bits/stdc++.h>
#define FT first
#define SC second
#define PB push_back
#define MP make_pair
#define REP(i, l, r) for(int i = (l); i <= (r); i++)
#define PER(i, r, l) for(int i = (r); i >= (l); i--)
#define FOR(i, n) for(int i = 0; i < (n); i++)
#define ROF(i, n) for(int i = (n) - 1; i >= 0; i--)
#define VEP(i, x) for(int i = 0; i < x.size(); i++)
#define DFOR(i, x, y) for(int i = hd[x], y = e[i].to; i; i = e[i].nxt, y = e[i].to)
#define MEM(a, b) memset(a, b, sizeof(a))
#define rint read<int>()
#define rll read<LL>()

using namespace std;
typedef long long LL;
typedef long double LD;
typedef pair<int, int> PI;
const int inf = 0x7fffffff;
const int MOD = 1000000007;

template <typename tn>
inline tn read(){
    char ch; tn f = 1;
    while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
    tn x = ch - '0';
    while (isdigit(ch = getchar())) x = x * 10 + ch - '0';
    return x * f;
}
template <typename tn> inline void cmax(tn &a, tn b){ if (a < b) a = b; }
template <typename tn> inline void cmin(tn &a, tn b){ if (a > b) a = b; }

const int N = 10000 + 5;
struct Edge{ int nxt, to; } e[N * 6];
struct Data{ int f, safe, danger; };
int color[N], x[N], y[N], z[N], sz[N], rt, tail, hd[N], tot, dep[N], S, T;
void add(int x, int y){ e[++tail] = (Edge){hd[x], y}, hd[x] = tail; }
Data dfs(int x, int c, int f){
    color[x] = c;
    Data cur;
    cur.safe = inf, cur.danger = -1, cur.f = 0;
    DFOR(i, x, y) if (y != f && color[y] != -2)
        if (!~color[y]){
            dep[y] = dep[x] + 1;
            Data now = dfs(y, c ^ 1, x); 
            cur.f += now.f;    if (cur.f > 1) { return cur;}
            cmin(cur.safe, now.safe), cmax(cur.danger, now.danger);
        } else if (dep[y] < dep[x]) if (color[x] ^ color[y]) cmin(cur.safe, dep[y]); else{
            cmax(cur.danger, dep[y]);
            if (!S) S = x, T = y;
            if (S != x && S != y) S = -1;
            if (T != x && T != y) T = -1;
        }
    if (cur.safe <= dep[x] && cur.danger >= dep[x]) cur.f = 100;
    else if (cur.danger >= dep[x]) cur.f++, cur.danger = -1;
    return cur;
}
int main(){
    freopen("river.in", "r", stdin);
    freopen("river.out", "w", stdout);
    int Cas = rint;
    while (Cas--){
        int n = rint, m = rint;
        MEM(sz, 0), MEM(hd, 0), tail = 0, rt = 0;
        REP(i, 1, m) sz[x[i] = rint]++, sz[y[i] = rint]++, sz[z[i] = rint]++;
        REP(i, 1, n) if (sz[i] == m) rt = i;
        if (rt){
            bool ans = 0;
            REP(i, 1, m){
                if (x[i] == rt) swap(x[i], z[i]);
                if (y[i] == rt) swap(y[i], z[i]);
                add(x[i], y[i]), add(y[i], x[i]);
            } 
            REP(i, 1, n) if (i != rt) { 
                MEM(color, -1), color[i] = -2;
                int flag = 1, tot = 0;
                REP(i, 1, n) if (color[i] == -1){
                    S = 0, T = 0;
                    int tmp = dfs(i, 0, 0).f;
                    if (S > 0 || T > 0) cmin(tmp, 1);
                    tot += tmp;
                    if (tot > 1) {flag = 0; break;}
                }
                if (flag) {ans = 1; break;}
            }
            if (ans) printf("yes
"); else printf("no
");
        } else printf("no
");
    }
}