BZOJ 2330. [SCOI2011]糖果 题解

题目描述

#2330. [SCOI2011]糖果

solve

可以把关系看成是一种约束，然后直接(SPFA)刷分差分约束就好了，貌似这道题卡(SPFA)看一波题解，发现反建图就好了

code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=100005,maxe=400005;
int lnk[maxn],nxt[maxe],son[maxe],w[maxe],dis[maxn],Q[maxn],cnt,vis[maxn],cir[maxn],N,K;
LL Ans;
inline void add_e(int x,int y,int z){
	son[++cnt]=y;w[cnt]=z;nxt[cnt]=lnk[x];lnk[x]=cnt;
}
inline int read(){
	int ret=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-f;ch=getchar();}
	while(ch<='9'&&ch>='0')ret=ret*10+ch-'0',ch=getchar();
	return ret*f;
}
int SPFA(){
	int hed=0,til=1;Q[til]=0;vis[0]=1;cir[0]=1;
	while(hed!=til){
		hed=(hed+1>=maxn?0:hed+1);vis[Q[hed]]=0;
		for(int j=lnk[Q[hed]];j;j=nxt[j]){
			if(dis[Q[hed]]+w[j]>dis[son[j]]){
				dis[son[j]]=dis[Q[hed]]+w[j];
				if(++cir[son[j]]>=N)return 0;
				if(!vis[son[j]])vis[son[j]]=1,Q[til=(til+1>=maxn?0:til+1)]=son[j];
			}
		}
	}
	return 1;
}
int main(){
	freopen("2330.in","r",stdin);
	freopen("2330.out","w",stdout);
	N=read();K=read();
	for(int i=1;i<=K;i++){
		int X=read(),A=read(),B=read();
		if(X==1){add_e(A,B,0);add_e(B,A,0);}
		else if(X==2){if(A==B){printf("-1
");return 0;}add_e(A,B,1);}
		else if(X==3){add_e(B,A,0);}
		else if(X==4){if(A==B){printf("-1
");return 0;}add_e(B,A,1);}
		else if(X==5){add_e(A,B,0);}
	}
	for(int i=N;i;i--)add_e(0,i,1);
	if(!SPFA()){printf("-1
");return 0;}
	for(int i=1;i<=N;i++)Ans+=dis[i];
	printf("%lld
",Ans);
	return 0;
}