Gym 102055B Balance of the Force

大意: $n$个骑士, 第$i$个骑士若加入光明阵营, 那么能力值$L_i$, 加入黑暗阵营, 能力值$D_i$. 给定$m$个限制$(u_i,v_i)$, 表示$u_i,v_i$不能在同一阵营. 求一种划分方案, 使得能力值最大值减最小值最小.

对于一个连通块, 如果不是二分图, 那么无解. 否则的话这个连通块最大值最小值只有两种情况, 枚举最大值, 求出最小值的最大可能值更新答案即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
 
const int N = 1e6+50;
int n,m,ok,vis[N],l[N],d[N],mi[N],ID[N],cur[N];
vector<int> g[N];
pii f[N],A,B;
 
void dfs(int x, int c) {
	vis[x] = c;
	if (c) {
		A.x = min(A.x,l[x]);
		A.y = max(A.y,l[x]);
		B.x = min(B.x,d[x]);
		B.y = max(B.y,d[x]);
	}
	else {
		A.x = min(A.x,d[x]);
		A.y = max(A.y,d[x]);
		B.x = min(B.x,l[x]);
		B.y = max(B.y,l[x]);
	}
	for (int y:g[x]) {
		if (vis[y]<0) dfs(y,c^1);
		else if (vis[y]==c) ok=0;
	}
}
 
void work() {
	scanf("%d%d",&n,&m);
	REP(i,1,n) vis[i]=-1,g[i].clear();
	REP(i,1,m) {
		int u, v;
		scanf("%d%d",&u,&v);
		g[u].pb(v),g[v].pb(u);
	}
	REP(i,1,n) scanf("%d%d",l+i,d+i);
	ok = 1;
	vector<pii> events;
	int cnt = 0;
	multiset<int> s;
	REP(i,1,n) if (vis[i]<0) {
		A = B = {1e9,0};
		dfs(i, 0);
		if (!ok) return puts("IMPOSSIBLE"),void();
		s.insert(cur[i]=-INF);
		ID[cnt]=i,mi[cnt]=A.x,events.pb(pii(A.y,cnt)),++cnt;
		ID[cnt]=i,mi[cnt]=B.x,events.pb(pii(B.y,cnt)),++cnt;
	}
	sort(events.begin(),events.end());
	int ans = 1e9;
	for (auto &p:events) {
		s.erase(s.find(cur[ID[p.y]]));
		cur[ID[p.y]] = max(cur[ID[p.y]], mi[p.y]);
		s.insert(cur[ID[p.y]]);
		ans = min(ans, p.x-*s.begin());
	}
	printf("%d\n", ans);
}
 
int main() {
	int t=rd();
	REP(i,1,t) {
		printf("Case %d: ",i);
		work();
	}
}