【Codeforces 536 D】Tavas in Kansas（最短路 / 博弈論 / DP）

傳送門

首先考慮對$S,T$求出到每個點最短距離後離散化
值域變成$O(n)$

對於每個點看做$a[dis_s][dis_t]=p_a$
那麼問題就變成在二維平面上做
設$f[0/1][x][y]$表示$先/後手，在\geq x,\geq y$的所有點中選的最大價值
注意這個要從後往前做
$cnt$表示後面點的數量，$s$表示權值和
那麼$f[0][x][y]=\max_{cnt[x'][y]>cnt[x][y]}(s[x][y]-f[1][x'][y])\\ f[1][x][y]=\max_{cnt[x][y']>cnt[x][y]}(s[x][y]-f[0][x][y'])$

記錄後綴$\min$即可做到$O(n^2)$

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define bg begin
#define fi first
#define se second
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
inline ll readll(){
	char ch=gc();
	ll res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
namespace Module{
cs int mod=998244353;
inline int add(int a,int b){return (a+b>=mod)?(a+b-mod):(a+b);}
inline int dec(int a,int b){return (a<b)?(a-b+mod):(a-b);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){a=(a+b>=mod)?(a+b-mod):(a+b);}
inline void Dec(int &a,int b){a=(a<b)?(a-b+mod):(a-b);}
inline void Mul(int a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(ll x){x%=mod,(x<0)&&(x+=mod);return x;}
}
cs int N=2005,M=100005;
typedef pair<ll,int> pli;
int n,m,x[N],y[N],S,T,p[N],ln,lm;
ll dis[N],b[N],s[N][N],mx[N][N],my[N][N],f[2][N][N];
bool vis[N],hv[N][N];
priority_queue<pli,vector<pli>,greater<pli> >q;
vector<pii> e[N];
inline void dijkstra(int *tmp,int &l,int st){
	memset(dis,127/3,sizeof(dis));
	memset(vis,0,sizeof(vis));
	dis[st]=0,q.push(pli(0,st));
	while(!q.empty()){
		int u=q.top().se;q.pop();
		if(vis[u])continue;vis[u]=1;
		for(pii &x:e[u])if(dis[x.fi]>dis[u]+x.se){
			dis[x.fi]=dis[u]+x.se;
			q.push(pli(dis[x.fi],x.fi));
		}
	}
	for(int i=1;i<=n;i++)b[i]=dis[i];
	sort(b+1,b+n+1);l=unique(b+1,b+n+1)-b-1;
	for(int i=1;i<=n;i++)tmp[i]=lower_bound(b+1,b+l+1,dis[i])-b;
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),m=read();
	S=read(),T=read();
	for(int i=1;i<=n;i++)p[i]=read();
	for(int i=1;i<=m;i++){
		int u=read(),v=read(),w=read();
		e[u].pb(pii(v,w)),e[v].pb(pii(u,w));
	}
	dijkstra(x,ln,S),dijkstra(y,lm,T);
	for(int i=1;i<=n;i++)hv[x[i]][y[i]]=1,s[x[i]][y[i]]+=p[i];
	for(int i=ln;i;i--)
	for(int j=lm;j;j--){
		s[i][j]+=s[i+1][j]+s[i][j+1]-s[i+1][j+1];
		if(!hv[i][j])
		mx[i][j]=min(i==ln?ln:mx[i+1][j],j==lm?ln:mx[i][j+1]),
		my[i][j]=min(i==ln?lm:my[i+1][j],j==lm?lm:my[i][j+1]);
		else mx[i][j]=i,my[i][j]=j;
		f[0][i][j]=s[i][j]-f[1][mx[i][j]+1][j];
		f[1][i][j]=s[i][j]-f[0][i][my[i][j]+1];
		if(i==1&&j==1)continue;
		chemn(f[0][i][j],f[0][i][j+1]),
		chemn(f[1][i][j],f[1][i+1][j]);
	}
	ll v1=f[0][1][1],v2=s[1][1]-f[0][1][1];
	if(v1<v2)puts("Cry");
	if(v1>v2)puts("Break a heart");
	if(v1==v2)puts("Flowers");
	return 0;
}