【bzoj2330】糖果 差分约束

AC通道：http://www.lydsy.com/JudgeOnline/problem.php?id=2330

【题解】

建图方法很简单（不会的去学差分约束），然后建立一个超级源S，由S向每个点连边，注意这里要倒着连，否则会超时（别问我为什么），然后跑最长路即可。

/*************  bzoj 2330  by chty  2016.11.16*************/#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;#define MAXN 1000100#define FILE "read"#define up(i,j,n)   for(ll i=j;i<=n;i++)#define down(i,j,n) for(ll i=j;i>=n;i--)namespace INIT{char buf[1<<15],*fs,*ft;inline char getc() {return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;}inline ll read(){ll x=0,f=1;  char ch=getc();while(!isdigit(ch))  {if(ch=='-')  f=-1;  ch=getc();}while(isdigit(ch))  {x=x*10+ch-'0';  ch=getc();}return x*f;}}using namespace INIT;struct node{ll y,next,v;}e[MAXN*2];ll n,m,ans,len,Link[MAXN],dis[MAXN],vis[MAXN],cnt[MAXN],q[MAXN*2];void insert(ll x,ll y,ll v) {e[++len].next=Link[x];Link[x]=len;e[len].y=y;e[len].v=v;}void init(){n=read();  m=read();up(i,1,m){ll flag=read(),x=read(),y=read();if(flag==1)  insert(x,y,0),insert(y,x,0);if(flag==2)  insert(x,y,1);if(flag==3)  insert(y,x,0);if(flag==4)  insert(y,x,1);if(flag==5)  insert(x,y,0);if(!(flag&1)&&x==y)  {puts("-1");  exit(0);}}down(i,n,1)  insert(0,i,1);}void spfa(){ll head=0,tail=1;memset(dis,-1,sizeof(dis));q[1]=0;  vis[0]=1;  dis[0]=0;while(++head<=tail){ll x=q[head];for(ll i=Link[x];i;i=e[i].next){if(dis[x]+e[i].v>dis[e[i].y]){dis[e[i].y]=dis[x]+e[i].v;if(!vis[e[i].y]){vis[e[i].y]=1;if(cnt[e[i].y]==n)  {puts("-1"); exit(0);}q[++tail]=e[i].y;cnt[e[i].y]++;}}}vis[x]=0;}}void solve(){spfa();up(i,1,n)  ans+=dis[i];printf("%lld\n",ans);}int main(){freopen(FILE".in","r",stdin);freopen(FILE".out","w",stdout);init();solve();return 0;}