POJ 3159(差分约束+dijkstra）

题目：输入n，m个关系。接下来m行每行A，B，C。表示B比A多出来的糖果不超过C个。

问：n号人最多比1号人多几个糖果

思路：差分约束关系，B-A<C可以看作由A做一条有向边到B，权值为C。问题变成了1到N的最短路

这题貌似只能用邻接表+优先队列，反正我平常都这么写的。这题同样也卡scanf

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
//const int inf = 0x6fffffff;

#define MAX INT_MAX
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
using namespace std;
int ver[151000],edge[151000],next[151000],head[151000],vis[151000],d[151000];
int n,m,tot;
void add(int x,int y,int z)
{
    ver[++tot] = y,edge[tot] = z,next[tot] = head[x] , head[x] = tot;
}
void djijiestra()
{
    priority_queue<pair<int,int> >que;
    memset(d,0x3f,sizeof(d));
    d[1] = 0;
    que.push(make_pair(0,1));
    while(que.size())
    {
        int x = que.top().second;que.pop();
        if(vis[x] == 1) continue;
        vis[x] = 1;
        for(int i=head[x];i;i=next[i])
        {
            int y = ver[i],z = edge[i];
            if(d[y] > d[x] + z)
            {
                d[y] = d[x] + z;
                que.push(make_pair(-d[y],y));
            }
        }
    }
}
int main()
{

    //ios::sync_with_stdio(false);
    //cin>>n>>m;
    scanf("%d %d",&n,&m);
    FOR(i,1,m)
    {
        int x,y,z;
        scanf("%d %d %d",&x,&y,&z);
        add(x,y,z);
    }
    djijiestra();
    printf("%d\n",d[n]);
}