2017 ACM/ICPC Asia Regional Shenyang Online spfa+最长路
transaction transaction transaction
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1496 Accepted Submission(s): 723
As we know, the price of this book was different in each city. It is ai yuan in it city. Kelukin will take taxi, whose price is 1yuan per km and this fare cannot be ignored.
There are n−1 roads connecting n cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get.
For each test case:
first line contains an integer n (2≤n≤100000) means the number of cities;
second line contains n numbers, the ith number means the prices in ith city; (1≤Price≤10000)
then follows n−1 lines, each contains three numbers x, y and z which means there exists a road between x and y, the distance is zkm (1≤z≤1000).
4
10 40 15 30
1 2 30
1 3 2
3 4 10
#include<bits/stdc++.h>
//#include<regex>
#define db double
#define ll long long
#define vec vector<ll>
#define Mt vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d\n",x)
#define pd(x) printf("%f\n",x)
#define pl(x) printf("%lld\n",x)
#define MP make_pair
#define PB push_back
#define fr(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=1e6+;
const int mod=1e9+;
const int MOD=mod-;
const db eps=1e-;
const db pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int a[N],d[N],vis[N]; struct P
{
int u,v,w;
P(int x,int y,int z):u(x),v(y),w(z){};
P(){};
};
vector<P> g[N],e;
queue<int> q;
void add(int x,int y,int z)
{
g[x].push_back(P(x,y,z));
}
void spfa(int n)
{
memset(d,, sizeof(d));
memset(vis,, sizeof(vis));
vis[]=;
q.push();
while(q.size())
{
int u=q.front();q.pop();
vis[u]=;
for(int i=;i<g[u].size();i++){
int v=g[u][i].v;
int w=g[u][i].w;
if(d[v]<d[u]+w){// get the maxmum
d[v]=d[u]+w;
if(!vis[v]){
vis[v]=;//push the new point
q.push(v);
}
}
}
}
pi(d[n+]);
}
int main()
{
int t;
ci(t);
while(t--)
{
int n;
ci(n);
for(int i=;i<=n;i++) g[i].clear();
for(int i=;i<=n;i++)
ci(a[i]),add(,i,a[i]),add(i,n+,-a[i]);
for(int i=;i<n;i++){
int x,y,z;
ci(x),ci(y),ci(z);
add(x,y,-z);
add(y,x,-z);
}
spfa(n); } }
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