【三中校内训练】净化

题解:SPFA裸题，首先计算出最短距离然后乱搞即可

#include<stdio.h>
#include<queue>
#include<math.h>
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(t) ((t)>0?(t):(-(t)))
using namespace std;
const int N=1000001;
struct edge{int next,to,len;} e[N];
queue<int> q;typedef long long ll;
int n,g[N],a[N],b[N],c[N],d[N],u[N],M,m;
ll dis[N],ans=0,t;bool inq[N];
inline void addedge(int x,int y,int z){e[++M]=(edge){g[x],y,z};g[x]=M;}
int main(){
    freopen("purify.in","r",stdin);freopen("purify.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1,x;i<=n;i++){
        scanf("%d",&x);
        if(x==1){q.push(i);dis[i]=0;inq[i]=true;}
        else dis[i]=(1ll<<50);
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
        if(c[i]==1) addedge(a[i],b[i],d[i]);
        else{addedge(a[i],b[i],d[i]);addedge(b[i],a[i],d[i]);}
    }
    while(!q.empty()){
        register int h=q.front();q.pop();inq[h]=false;
        for(register int i=g[h];i;i=e[i].next) if(dis[e[i].to]>dis[h]+e[i].len){
            dis[e[i].to]=dis[h]+e[i].len;
            if(!inq[e[i].to]){q.push(e[i].to);inq[e[i].to]=true;}
        }
    }
    for(register int i=1,x,y;i<=m;i++){
        x=a[i];y=b[i];
        if(c[i]==1) ans=max(ans,dis[x]+d[i]);
        else{
            t=dis[x]-dis[y];
            if(abs(t)>d[i]) ans=max(ans,max(dis[x],dis[y]));
            else ans=max(ans,ceil(max((-t+d[i])/2.0+dis[x],(t+d[i])/2.0+dis[y])));
        }
    }
    printf("%I64d\n",ans);
    return 0;
}

【我保证以后不恶意压行了！】