P1073 最优贸易 建立分层图 + spfa

P1073 最优贸易：https://www.luogu.org/problemnew/show/P1073

题意：

　　　有n个城市，每个城市对A商品有不同的定价，问从1号城市走到n号城市可以最多赚多少差价。（旅游为主，赚钱为辅，所以买入和卖出只进行一次。

思路：

　　建一个有三层的图，三层都是相同的普通的城市路线，第一层向第二层连从第i个城市买入商品的花费，第二层向第三层连从第i个城市卖出商品的所得。从1 向 第一层的终点 ，向第三层的终点跑一遍最大路就行了。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1000000007;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/
          
            const int maxn = 1e5+9;

            vector<pii>mp[maxn*3];
            int a[maxn];
            int n,m;
            void addedge(int s,int t){
                mp[s].pb(pii(t, 0));
                mp[s + n].pb(pii(t + n,0));
                mp[s + 2 * n].pb(pii(t + 2 * n,0));

                mp[s].pb(pii(t + n, -a[s]));
                mp[s + n].pb(pii(t + 2 * n, a[s]));
            }

            int dis[maxn * 3];
            bool vis[maxn * 3];

            int spfa(int s,int t){
               
                memset(dis, -inf,sizeof(dis));
                dis[s] = 0;
                queue<int>que;  que.push(s);
                vis[s] = true;
                while(!que.empty()){
                    int u = que.front(); que.pop();
                    vis[u] = false;
                    for(int i=0; i<mp[u].size(); i++){
                        int v = mp[u][i].fi,w = mp[u][i].se;
                        if(dis[v] < dis[u] + w){
                            dis[v] = dis[u] + w;
                            if(vis[v] == false){
                                vis[v] = true;
                                que.push(v);
                            }
                        }
                    }
                }
                return dis[t];
            }
int main(){
            scanf("%d%d", &n, &m);
            int t = 3 * n + 1;
            for(int i=1; i<=n; i++) scanf("%d", &a[i]);
            for(int i=1; i<=m; i++){
                int u,v,c;
                scanf("%d%d%d", &u, &v, &c);
                if(c == 1) {
                    addedge(u,v);
                }
                else {
                    addedge(u,v);
                    addedge(v,u);
                }
            }
           
            // addedge(3*n, t);
            mp[3*n].pb(pii(t,0));
            mp[n].pb(pii(t, 0));
            printf("%d\n", spfa(1, t));
            return 0;
}

标签：pii,ch,int,分层,spfa,que,include,P1073,define
来源： https://www.cnblogs.com/ckxkexing/p/10351183.html