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.

 

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#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();     } }