transaction transaction transaction
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1244 Accepted Submission(s): 595
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).
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
#define N 100005
#define mem(a,x) memset(a,x,sizeof(a))
struct data
{
int to;
int cost;
};
int n;
int dis[N];
int vis[N];
int Point[N];
vector<data>v[N];
void init()
{
for(int i=0; i<N; i++)
v[i].clear();
}
int add(int a,int b,int c)
{
data tmp;
tmp.cost=c;
tmp.to=b;
v[a].push_back(tmp);
}
void spfa(int u)
{
memset(vis,0,sizeof(vis));
memset(dis,-inf,sizeof(dis));
queue<int>q;
vis[u]=1;
q.push(u);
dis[u]=0;
while(!q.empty())
{
u=q.front();
q.pop();
vis[u]=0;
int len=v[u].size();
for(int i=0; i<len; i++)
{
int to=v[u][i].to;
int cost=v[u][i].cost;
int tmp=dis[u]+cost;
if(tmp>dis[to])
{
dis[to]=tmp;
if(!vis[to])
{
vis[to]=1;
q.push(to);
}
// printf("--%d\n",u);
}
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
init();
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d",&Point[i]);
}
Point[0]=Point[n+1]=0;
for(int i=1; i<=n; i++)
{
add(0,i,-2);///爲新建源點與各點的邊賦值
add(i,n+1,2);///爲新建匯點與各點的邊賦值 距離隨便填了,,只要對稱就好
}
int x,y,c;
for(int i=1; i<n; i++)
{
scanf("%d%d%d",&x,&y,&c);
add(x,y,Point[y]-Point[x]-c);
add(y,x,Point[x]-Point[y]-c);
}
spfa(0);
//printf("---------\n");
/*for(int i=0;i<=n+1;i++)
{
printf("%d ",dis[i]);
}
printf("\n");*/
printf("%d\n",dis[n+1]);
}
}