• 2017-2018 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2017)


    A. Airport Coffee

    设$f_i$表示考虑前$i$个咖啡厅,且在$i$处买咖啡的最小时间,通过单调队列优化转移。

    时间复杂度$O(n)$。

    #include<cstdio>
    const int N=500010;
    typedef long long ll;
    const double inf=1e20;
    ll L,A,B,T,R,lim0,lim1,a[N];
    double f[N],val[N];
    double dp0=inf;
    int id0;
    int pre[N];
    int i,j,k;
    int h=1,t,q[N];
    double ans;
    int fin;
    int cof[N],cnt,n;
    int main(){
    	scanf("%lld%lld%lld%lld%lld",&L,&A,&B,&T,&R);
    	lim0=A*T+R*B;
    	lim1=T*A;
    	scanf("%d",&n);
    	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
    	ans=1.0*L/A;
    	for(i=j=k=1;i<=n;i++){
    		f[i]=1.0*a[i]/A;
    		
    		while(j<i&&a[i]-a[j]>=lim0){
    			double now=f[j]-1.0*a[j]/A;
    			if(now<dp0){
    				dp0=now;
    				id0=j;
    			}
    			j++;
    		}
    		if(id0){
    			double now=dp0+1.0*(a[i]-lim0)/A+T+R;
    			if(now<f[i]){
    				f[i]=now;
    				pre[i]=id0;
    			}
    		}
    		
    		while(k<i&&a[i]-a[k]>=lim1){
    			val[k]=f[k]-1.0*a[k]/B;
    			while(h<=t&&val[q[t]]>val[k])t--;
    			q[++t]=k;
    			k++;
    		}
    		while(h<=t&&a[i]-a[q[h]]>=lim0)h++;
    		if(h<=t){
    			double now=val[q[h]]+1.0*(a[i]-lim1)/B+T;
    			if(now<f[i]){
    				f[i]=now;
    				pre[i]=q[h];
    			}
    		}
    		double ret=f[i];
    		ll d=L-a[i];
    		if(d<=lim1){
    			ret+=1.0*d/A;
    		}else if(d<=lim0){
    			ret+=1.0*(d-lim1)/B+T;
    		}else{
    			ret+=1.0*(d-lim0)/A+T+R;
    		}
    		if(ret<ans){
    			ans=ret;
    			fin=i;
    		}
    	}
    	while(fin){
    		cof[++cnt]=fin;
    		fin=pre[fin];
    	}
    	printf("%d
    ",cnt);
    	for(i=cnt;i;i--)printf("%d ",cof[i]-1);
    }
    //0 .. n-1
    

      

    B. Best Relay Team

    按题意模拟即可。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define MC(x, y) memcpy(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1010, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int l, r;
    struct A
    {
    	string s;
    	double fi, se;
    }a[N], b[N];
    int n;
    string ss[5];
    
    bool cmp(A a, A b)
    {
    	return a.se < b.se;
    }
    int main()
    {
    	scanf("%d", &n);
    	for(int i = 1; i <= n; i ++){
    		cin >> a[i].s;
    		scanf("%lf%lf", &a[i].fi, &a[i].se);
    		a[i + n] = a[i];
    	}
    	double ans = 1e9;
    	for(int i = 1; i <= n; i ++){
    		MC(b, a);
    		sort(b + i + 1, b + n + i, cmp);
    		double tmp = b[i].fi + b[i + 1].se + b[i + 2].se + b[i + 3].se; 
    		if(tmp < ans){
    			ans = tmp;
    			ss[1] = b[i].s;
    			ss[2] = b[i + 1].s;
    			ss[3] = b[i + 2].s;
    			ss[4] = b[i + 3].s;
    		}
    	}
    	printf("%.2f
    ", ans);
    	for(int i = 1; i <= 4; i ++){
    		//printf("%s
    ", ss[i]);	
    		cout << ss[i] << endl;
    	}
    	scanf("%d", &n);
    	
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    C. Compass Card Sales

    按题意模拟即可。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n;
    struct A
    {
    	int v[3];
    	int id;
    	int val;
    }a[N];// keep a not modified
    struct B
    {
    	int id;
    	int val;
    	bool operator < (const B & b)const
    	{
    		if(val != b.val)return val < b.val;
    		return id > b.id;
    	}
    };
    set<int>sot[3][720];
    set<B>b;
    //use the a[o]
    int getval(int o)
    {
    	int val = 0;
    	for(int i = 0; i < 3; ++i)
    	{
    		int pos = a[o].v[i];
    		if(sot[i][pos].size() >= 2)
    		{
    			continue;
    		}
    		int pre = pos + 360;
    		do
    		{		
    			--pre;
    		}while(!sot[i][pre].size());
    		int suf = pos;
    		do
    		{
    			++suf;
    		}while(!sot[i][suf].size());
    		val += (pos + 360 - pre) + (suf - pos);
    	}
    	//printf("o = %d a[o].id = %d a[o].val = %d
    ", o, a[o].id, val);
    	return val;
    }
    map<int, int>mop;
    int main()
    {
    	while(~scanf("%d",&n))
    	{
    		mop.clear();
    		for(int i = 0; i < 3; ++i)
    		{
    			for(int j = 0; j < 720; ++j)
    			{
    				sot[i][j].clear();
    			}
    		}
    		b.clear();
    		for(int i = 1; i <= n; ++i)
    		{
    			for(int j = 0; j < 3; ++j)
    			{
    				scanf("%d", &a[i].v[j]);
    			}
    			scanf("%d", &a[i].id);
    			mop[a[i].id] = i;
    			for(int j = 0; j < 3; ++j)
    			{
    				int p = a[i].v[j];
    				int o = a[i].id;
    				sot[j][p].insert(o);
    				sot[j][p + 360].insert(o);
    			}
    		}
    		//a index is a input order
    		for(int i = 1; i <= n; ++i)
    		{
    			a[i].val = getval(i);
    			b.insert({a[i].id, a[i].val});
    		}
    		while(n--)
    		{
    			int id = b.begin()->id;
    			int o = mop[id];
    			printf("%d
    ", id);
    			b.erase({id, a[o].val});
    			for(int i = 0; i < 3; ++i)
    			{
    				int pos = a[o].v[i];
    				sot[i][pos].erase(id);
    				sot[i][pos + 360].erase(id);
    			}
    			
    			for(int i = 0; i < 3; ++i)
    			{
    				int pos = a[o].v[i];
    				if(sot[i][pos].size() == 1)
    				{
    					int rst = mop[*sot[i][pos].begin()];
    					b.erase({a[rst].id, a[rst].val});
    					a[rst].val = getval(rst);
    					b.insert({a[rst].id, a[rst].val});
    				}
    				else if(sot[i][pos].size() == 0 && n > 1)
    				{		
    					int pos = a[o].v[i];
    					int pre = pos + 360;
    					do
    					{		
    						--pre;
    					}while(!sot[i][pre].size());
    					if(sot[i][pre].size() == 1)
    					{
    						int rst = mop[*sot[i][pre].begin()];
    						b.erase({a[rst].id, a[rst].val});
    						a[rst].val = getval(rst);
    						b.insert({a[rst].id, a[rst].val});
    					}
    					int suf = pos;
    					do
    					{
    						++suf;
    					}while(!sot[i][suf].size());
    					if(sot[i][suf].size() == 1)
    					{
    						int rst = mop[*sot[i][suf].begin()];
    						b.erase({a[rst].id, a[rst].val});
    						a[rst].val = getval(rst);
    						b.insert({a[rst].id, a[rst].val});
    					}
    				}
    			}
    		}
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    D. Distinctive Character

    设$f_S$表示到$S$的最大相似度,显然对于输入的$n$个串$a_i$,有$f_{a_i}=k$。

    对于$f_S=k$,将$S$修改一位,可以得到$f_{S'}=k-1$,BFS求出所有$S$即可。

    时间复杂度$O(k2^k)$。

    #include<cstdio>
    int n,m,i,h,t,f[1<<20],q[2222222],x,y;
    char s[100];
    inline void ext(int x,int y){if(f[x]<0)f[q[++t]=x]=y;}
    int main(){
    	scanf("%d%d",&n,&m);
    	for(i=0;i<1<<m;i++)f[i]=-1;
    	while(n--){
    		scanf("%s",s);
    		int t=0;
    		for(i=0;i<m;i++)t=t*2+s[i]-'0';
    		f[t]=m;
    	}
    	h=1;
    	for(i=0;i<1<<m;i++)if(~f[i])q[++t]=i;
    	while(h<=t){
    		x=q[h++];
    		y=f[x]-1;
    		for(i=0;i<m;i++)ext(x^(1<<i),y);
    	}
    	for(i=m-1;~i;i--)printf("%d",x>>i&1);
    }
    

      

    E. Emptying the Baltic

    令两点间的边权为较大的高度,求出最小生成树,则每个点的实际水位为起点到它路径上点权的最大值。

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int N=550,M=N*N;
    int n,m,i,j,x,y,tot,id[N][N],a[M],f[M];
    int ce;
    int v[M*2],nxt[M*2],ed,g[M];
    long long ans;
    struct E{int x,y,z;E(){}E(int _x,int _y,int _z){x=_x,y=_y,z=_z;}}e[M*9];
    inline bool cmp(const E&a,const E&b){return a.z<b.z;}
    int F(int x){return f[x]==x?x:f[x]=F(f[x]);}
    inline void add(int x,int y){
    	v[++ed]=y;nxt[ed]=g[x];g[x]=ed;
    }
    void dfs(int x,int y,int z){
    	z=max(z,a[x]);
    	if(z<0)ans-=z;
    	for(int i=g[x];i;i=nxt[i])if(v[i]!=y)dfs(v[i],x,z);
    }
    int main(){
    	scanf("%d%d",&n,&m);
    	for(i=1;i<=n;i++)for(j=1;j<=m;j++){
    		id[i][j]=++tot;
    		scanf("%d",&a[tot]);
    	}
    	for(i=1;i<=n;i++)for(j=1;j<=m;j++)for(x=-1;x<=1;x++)for(y=-1;y<=1;y++){
    		int nx=i+x,ny=j+y;
    		if(nx<1||nx>n||ny<1||ny>m)continue;
    		e[++ce]=E(id[i][j],id[nx][ny],max(a[id[i][j]],a[id[nx][ny]]));
    	}
    	sort(e+1,e+ce+1,cmp);
    	for(i=1;i<=tot;i++)f[i]=i;
    	for(i=1;i<=ce;i++)if(F(e[i].x)!=F(e[i].y)){
    		f[f[e[i].x]]=f[e[i].y];
    		add(e[i].x,e[i].y);
    		add(e[i].y,e[i].x);
    	}
    	scanf("%d%d",&i,&j);
    	dfs(id[i][j],0,-10000000);
    	printf("%lld",ans);
    }
    

      

    F. Fractal Tree

    留坑。

    G. Galactic Collegiate Programming Contest

    用一个set维护所有严格比$1$好的队伍,若别人过题则试情况插入set,若$1$过题则将set中最差的若干队伍删除。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n, m;
    struct A
    {
    	int g, t;
    	bool operator < (const A &b)const
    	{
    		if(g != b.g)return g > b.g;
    		return t < b.t;
    	}
    }a[N];
    multiset<A>sot;
    int main()
    {
    	while(~scanf("%d%d",&n, &m))
    	{
    		for(int i = 1; i <= n; ++i)
    		{
    			a[i].g = a[i].t = 0;
    		}
    		for(int i = 1; i <= m; ++i)
    		{
    			int o, p;
    			scanf("%d%d", &o, &p);
    			if(o == 1)
    			{
    				a[1].g += 1;
    				a[1].t += p;
    			}
    			else
    			{
    				if(a[o] < a[1])
    				{
    					sot.erase(sot.find(a[o]));
    				}
    				a[o].g += 1;
    				a[o].t += p;
    				sot.insert(a[o]);
    			}
    			while(!sot.empty() && !(*--sot.end() < a[1]))
    			{
    				sot.erase(--sot.end());
    			}
    			printf("%d
    ", sot.size() + 1);
    		}
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    H. Hubtown

    极角排序后求出每个人最近的$1$到$2$条火车线路,然后求最大流即可。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long ll;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 4e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n, m;
    int ST, ED, ID;
    int first[N], w[M], cap[M], nxt[M];
    void ins(int x, int y, int cap_)
    {
    	w[++ID] = y;
    	cap[ID] = cap_;
    	nxt[ID] = first[x];
    	first[x] = ID;
    	w[++ID] = x;
    	cap[ID] = 0;
    	nxt[ID] = first[y];
    	first[y] = ID;
    }
    int d[N];
    bool bfs()
    {
    	MS(d, -1); d[ST] = 0;
    	queue<int>q; q.push(ST);
    	while(!q.empty())
    	{
    		int x = q.front(); q.pop();
    		for(int z = first[x]; z; z = nxt[z])if(cap[z])
    		{
    			int y = w[z];
    			if(d[y] == -1)
    			{
    				d[y] = d[x] + 1;
    				if(y == ED)return 1;
    				q.push(y);
    			}
    		}
    	}
    	return 0;
    }
    int dfs(int x, int all)
    {
    	if(x == ED)return all;
    	int use = 0;
    	for(int z = first[x]; z; z = nxt[z])if(cap[z])
    	{
    		int y = w[z];
    		if(d[y] == d[x] + 1)
    		{
    			int tmp = dfs(y, min(cap[z], all - use));
    			cap[z] -= tmp;
    			cap[z ^ 1] += tmp;
    			use += tmp;
    			if(use == all)break;
    		}
    	}
    	if(use == 0)d[x] = -1;
    	return use;
    }
    int dinic()
    {
    	int tmp = 0;
    	while(bfs())tmp += dfs(ST, inf);
    	return tmp;
    }
    struct A
    {
    	int x, y, o;
    }a[N];
    struct B
    {
    	int x, y, o,c;
    }b[N];
    struct E{
    	int x,y,t;
    	int sgn;
    	E(){}
    	E(int _x,int _y,int _t){
    		x=_x,y=_y,t=_t;
    		if(!x)sgn=y>0;
    		else sgn=x>0;
    	}
    }e[400010];
    inline bool cmpe(const E&a,const E&b){
    	if(a.sgn!=b.sgn)return a.sgn<b.sgn;
    	int t=a.x*b.y-a.y*b.x;
    	if(t)return t<0;
    	return a.t<b.t;
    }
    int pre[400010],suf[400010];
    int gcd(int a,int b){return b?gcd(b,a%b):a;}
    inline bool point_on_line(const A&a,const B&b){
    	int d1=gcd(abs(a.x),abs(a.y)),d2=gcd(abs(b.x),abs(b.y));
    	return (a.x/d1==b.x/d2)&&(a.y/d1==b.y/d2);
    }
    struct Point{
    	ll x,y;
    	Point(){}
    	Point(ll _x,ll _y){x=_x,y=_y;}
    };
    inline ll cross(const Point&a,const Point&b){return a.x*b.y-a.y*b.x;}
    inline ll sig(ll x){
    	if(x==0)return 0;
    	return x>0?1:-1;
    }
    typedef long double ld;
    
    
    const ld epsss=1e-10;
    
    struct Pointd{
    	ld x,y;
    	Pointd(){}
    	Pointd(ld _x,ld _y){x=_x,y=_y;}
    };
    inline ld crossd(const Pointd&a,const Pointd&b){return a.x*b.y-a.y*b.x;}
    inline ll sigd(ld x){
    	if(fabs(x)<epsss)return 0;
    	return x>0?1:-1;
    }
    
    inline int distance_cmp(const A&_a,const B&_b,const B&_c){
    	Point a(_a.x,_a.y);
    	Point b(_b.x,_b.y);
    	Point c(_c.x,_c.y);
    	Point d;
    	if(!cross(b,c)){
    		d=Point(-b.y,b.x);
    		if(!cross(a,d))return 0;
    		if(sig(cross(d,a))==sig(cross(d,b)))return -1;
    		return 1;
    	}
    	ld L=sqrt(b.x*b.x+b.y*b.y);
    	ld R=sqrt(c.x*c.x+c.y*c.y);
    	Pointd aa(a.x,a.y);
    	Pointd bb(b.x,b.y);
    	Pointd cc(c.x,c.y);
    	Pointd dd(d.x,d.y);
    	bb.x*=R;
    	bb.y*=R;
    	cc.x*=L;
    	cc.y*=L;
    	dd=Pointd(bb.x+cc.x,bb.y+cc.y);
    	if(!sigd(crossd(aa,dd)))return 0;
    	if(sigd(crossd(dd,aa))==sigd(crossd(dd,bb)))return -1;
    	return 1;
    }
    int main()
    {
    	while(~scanf("%d%d",&n, &m))
    	{
    		ST = 0;
    		ED = n + m + 1;
    		ID = 1; MS(first, 0);
    		for(int i = 1; i <= n; ++i)
    		{
    			scanf("%d%d", &a[i].x, &a[i].y);
    			a[i].o = i;
    		}
    		//sort a?
    		
    		for(int i = 1; i <= m; ++i)
    		{
    			scanf("%d%d%d", &b[i].x, &b[i].y,&b[i].c);
    			b[i].o = i;
    		}
    		//sort b?
    		int ce=0;
    		for(int i=1;i<=n;i++){
    			e[++ce]=E(a[i].x,a[i].y,i);
    		}
    		for(int i=1;i<=m;i++){
    			e[++ce]=E(b[i].x,b[i].y,-i);
    		}
    		sort(e+1,e+ce+1,cmpe);
    		
    		pre[0]=0;
    		for(int i=1;i<=ce;i++)if(e[i].t<0)pre[0]=-e[i].t;
    		for(int i=1;i<=ce;i++){
    			pre[i]=pre[i-1];
    			if(e[i].t<0)pre[i]=-e[i].t;
    		}
    		
    		suf[ce+1]=0;
    		for(int i=ce;i;i--)if(e[i].t<0)suf[ce+1]=-e[i].t;
    		for(int i=ce;i;i--){
    			suf[i]=suf[i+1];
    			if(e[i].t<0)suf[i]=-e[i].t;
    		}
    		
    		for(int i=1;i<=ce;i++)if(e[i].t>0){
    			int x=e[i].t;
    			int L=pre[i],R=suf[i];
    			if(L==R){
    				ins(x,L+n,1);
    				continue;
    			}
    			if(point_on_line(a[x],b[L])){
    				ins(x,L+n,1);
    				continue;
    			}
    			if(point_on_line(a[x],b[R])){
    				ins(x,R+n,1);
    				continue;
    			}
    			int t=distance_cmp(a[x],b[L],b[R]);
    			if(t<=0)ins(x,L+n,1);
    			if(t>=0)ins(x,R+n,1);
    		}
    		for(int i = 1; i <= n; ++i)
    		{
    			ins(ST, i, 1);
    		}
    		for(int i = 1; i <= m; ++i)
    		{
    			ins(n + i, ED, b[i].c);
    		}
    		
    		//addedge: &&maybe use a little greedy method
    		
    		printf("%d
    ",dinic());
    		for(int i = 1; i <= n; ++i)
    		{
    			for(int z = first[i]; z; z = nxt[z])
    			{
    				int y = w[z];
    				if(y > n && cap[z] == 0)
    				{
    					printf("%d %d
    ", i - 1, y - n - 1);
    				}
    			}
    		}
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    I. Import Spaghetti

    枚举起点,BFS求最小环。

    #include<cstdio>
    #include<map>
    #include<cstring>
    #include<iostream>
    #include<sstream>
    using namespace std;
    const int N=555;
    int n,i;
    string name[N];
    bool g[N][N];
    map<string,int>id;
    int ans=N,finS,finT;
    int d[N],f[N],q[N],h,t;
    inline int ask(string t){
    	string x="";
    	for(int i=0;i<t.size();i++)if(t[i]>='a'&&t[i]<='z')x.push_back(t[i]);
    	return id[x];
    }
    inline void ext(int x,int y,int z){
    	if(d[x]<0){
    		q[++t]=x;
    		d[x]=y;
    		f[x]=z;
    	}
    }
    void bfs(int S){
    	int i,j,x;
    	for(i=1;i<=n;i++)d[i]=-1;
    	h=1,t=0;
    	ext(S,0,0);
    	while(h<=t){
    		x=q[h++];
    		int y=d[x]+1;
    		for(i=1;i<=n;i++)if(g[x][i])ext(i,y,x);
    	}
    	for(i=1;i<=n;i++)if(d[i]>0&&g[i][S]){
    		int now=d[i]+1;
    		if(now<ans)ans=now,finS=S,finT=i;
    	}
    }
    int main(){
    	scanf("%d",&n);
    	for(i=1;i<=n;i++){
    		cin>>name[i];
    		id[name[i]]=i;
    	}
    	for(i=1;i<=n;i++){
    		string tmp;int k;
    		cin>>tmp>>k;
    		char ss[2]; gets(ss);
    		while(k--)
    		{
    			string s;
    			getline(cin, s);
    			stringstream cinn(s);
    			int flag=1;
    			while(cinn>>s)
    			{
    				if(flag){flag=0;continue;}
    				int y=ask(s);
    				g[y][i]=1;
    				//printf("%d %d
    ",i,y);
    			}
    		}
    	}
    	for(i=1;i<=n;i++)if(g[i][i]){
    		cout<<name[i]<<endl;
    		return 0;
    	}
    	for(i=1;i<=n;i++)bfs(i);
    	if(ans>=N){
    		puts("SHIP IT");
    		return 0;
    	}
    	//printf("%d %d %d
    ",ans,finS,finT);
    	bfs(finS);
    	i=finT;
    	while(i){
    		cout<<name[i]<<" ";
    		i=f[i];
    	}
    }
    /*
    4
    a b c d
    a 1
    import d, b, c
    b 2
    import d
    import c
    c 1
    import c
    d 0
    
    
    5
    classa classb myfilec execd libe
    classa 2
    import classb
    import myfilec, libe
    classb 1
    import execd
    myfilec 1
    import libe
    execd 1
    import libe
    libe 0
    
    
    5
    classa classb myfilec execd libe
    classa 2
    import classb
    import myfilec, libe
    classb 1
    import execd
    myfilec 1
    import libe
    execd 1
    import libe, classa
    libe 0
    */
    

      

    J. Judging Moose

    按题意模拟即可。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int l, r;
    int main()
    {
    	while(~scanf("%d%d",&l, &r))
    	{
    		if(l + r == 0)
    		{
    			puts("Not a moose");
    		}
    		else if(l == r)
    		{
    			printf("Even %d
    ", l + r);
    		}
    		else 
    		{
    			printf("Odd %d
    ", max(l, r) * 2);
    		}
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    K. Kayaking Trip

    二分答案,贪心配对检验。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n, m;
    int g[3], gg[3];
    int s[3];
    int c[N];
    bool check(int aim)
    {
    	gg[0] = g[0];
    	gg[1] = g[1];
    	gg[2] = g[2];
    	for(int i = 1; i <= m; ++i)
    	{
    		int need = aim / c[i] + (aim % c[i] > 0);
    		int jj = -1;
    		int kk = -1;
    		for(int j = 0; j < 3; ++j)
    		{
    			for(int k = j; k < 3; ++k)
    			{
    				int num = 1;
    				if(j == k)num = 2;
    				if(gg[j] >= num && gg[k] >= num && s[j] + s[k] >= need)
    				{					
    					if(jj == -1 || s[j] + s[k] < s[jj] + s[kk])
    					{
    						jj = j;
    						kk = k;
    					}
    				}
    			}
    		}
    		if(jj == -1)return 0;
    		gg[jj] -= 1;
    		gg[kk] -= 1;
    		//
    		//printf("after ope %d: %d %d %d
    ", i, gg[0], gg[1], gg[2]);
    		//
    	}
    	return 1;
    }
    int main()
    {
    	while(~scanf("%d%d%d",&g[0], &g[1], &g[2]))
    	{
    		scanf("%d%d%d",&s[0], &s[1], &s[2]);
    		m = (g[0] + g[1] + g[2]) / 2;
    		for(int i = 1; i <= m; ++i)scanf("%d", &c[i]);
    		sort(c + 1, c + m + 1);
    		int l = c[1] * (s[0] + s[0]);
    		int r = c[m] * (s[2] + s[2]);
    		//
    		//l = r = 505;
    		//printf("l == %d r == %d
    ", l, r);
    		//
    		int ans = -1;
    		while(l <= r)
    		{
    			int mid = (l + r + 1) / 2;
    			if(check(mid))
    			{
    				ans = mid;
    				l = mid + 1;
    			}
    			else
    			{
    				r = mid - 1;
    			}
    		}
    		printf("%d
    ", ans);
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

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  • 原文地址:https://www.cnblogs.com/clrs97/p/7923485.html
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