题解 P1073 【最优贸易】

题目

我的思路是bfs/dfs+spfa

第一遍判断每个点能否到达n

第二遍找出：每个点，能到达这个点的所有点中价格最小是多少，在这个点卖掉最多能赚多少钱

上代码：（前面一大坨自定义不要管^_^

#include<bits/stdc++.h>
namespace ZDY{
#define fre freopen("test.in","r",stdin)
#define ll long long
#define db double
#define Fur(i,x,y) for(int i=x;i<=y;i++)
#define fur(i,x,y) for(i=x;i<=y;i++)
#define Fdr(i,x,y) for(int i=x;i>=y;i--)
#define in2(x,y) in(x),in(y)
#define in3(x,y,z) in2(x,y),in(z)
#define in4(a,b,c,d) in2(a,b);in2(c,d)
#define clr(x,y) memset(x,y,sizeof(x))
#define cpy(x,y) memcpy(x,y,sizeof(x))
#define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
#define inf 2147483647
}
namespace FAST{
#define pob (fwrite(fob::b,sizeof(char),fob::f-fob::b,stdout),fob::f=fob::b,0)
#define pc(x) (*(fob::f++)=(x),(fob::f==fob::g)?pob:0)
#define gc ((*fib::f)?(*(fib ::f++)):(fgets(fib::b,sizeof(fib::b),stdin)?(fib::f=fib::b,*(fib::f++)):-1))
namespace fib{char b[300000]= {},*f=b;}namespace fob{char b[300000]= {},*f=b,*g=b+300000-2;}struct foce{~foce(){pob;fflush(stdout);}} _foce;namespace ib{char b[100];}template<class T>inline void in(T &x){x=0;char c;bool f=0;while((c=gc)>'9'||c<'0')if(c=='-')f=!f;x=c-48;while((c=gc)<='9'&&c>='0')x=x*10+c-48;if(f)x=-x;}template<class T>inline void out(T x){if(x==0){pc(48);return;}if(x<0){pc('-');x=-x;}char *s=ib::b;while(x) *(++s)=x%10,x/=10;while(s!=ib::b) pc((*(s--))+48);}template<class T>inline void outn(T x){out(x);pc('
');}inline char sc(){char c=gc;while(!(('A'<=c&&c<='Z')||('a'<=c&&c<='z')||('0'<=c&&c<='9')))c=gc;return c;}
template<class T>inline T ABS(T x){return x>0?x:-x;}
template<class T>inline T MAX(T x,T y){return x>y?x:y;}
template<class T>inline T MIN(T x,T y){return x<y?x:y;}
template<class T>inline T GCD(T x,T y){return y?GCD(y,x%y):x;}
template<class T>inline void swap(T x,T y){T t=x;y=t;x=y;}
}using namespace FAST;using namespace ZDY;using namespace std;
#define N 100001
int val[N],head[N],Head[N],cnt=0,n,m,ans=0,q[N*3],f[N];
bool b[N],v[N];
struct edge{int nxt,to;}e[1000001],E[1000001];
inline void add(int x,int y){
    e[++cnt].to=y;E[cnt].to=x;
    e[cnt].nxt=head[x];E[cnt].nxt=Head[y];
    head[x]=Head[y]=cnt;
}
inline void bfs(){
    int h=0,t=1,x;
    q[0]=n;b[n]=1;
    while(h<t){
        x=q[h++];
        for(int i=Head[x],to;to=E[i].to,i;i=E[i].nxt)
        if(!b[to])b[to]=1,q[t++]=to;
    }
}
inline void bfs2(){
    int h=0,t=1,x;
    q[0]=1;
    while(h<t){
        x=q[h++];ans=MAX(ans,val[x]-f[x]);
        fl(i,x)
        if((f[x]<f[to]||!v[to])&&b[to])f[to]=MIN(f[x],f[to]),q[t++]=to;
    }
}
int main(){
    fre;
    in2(n,m);
    Fur(i,1,n)in(val[i]),f[i]=val[i];
    int x,y,z;
    Fur(i,1,m){
        in3(x,y,z);
        add(x,y);if(z==2)add(y,x);
    }
    bfs();
    bfs2();
    outn(ans);
}