POJ 3159 Candies(差分约束+dijkstra+heap)

Candies
Time Limit: 1500MS Memory Limit: 131072K
Total Submissions: 30002 Accepted: 8320
Description

During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.

snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?

Input

The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.

Output

Output one line with only the largest difference desired. The difference is guaranteed to be finite.

Sample Input

2 2
1 2 5
2 1 4
Sample Output

5
Hint

32-bit signed integer type is capable of doing all arithmetic.

题意：
给出n个点和m个关系，分别是A，B，C
意思是B-A<=C,求最后n能比1大多少。

解题思路：
给出A和B的关系是B-A<=C,那么就有dis[B]-dis[A] <= mp[A][B].
因为是求N的最大值，那么就使得
if(dis[B]>dis[A]+mp[A][B]) dis[B] = dis[A] + mp[A][B].
这里要用到优先队列优化dijkstra，优先队列里存放的是1到当前点的最短距离。
存图方式用的是链式前向星。

AC代码：

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <queue>#include <vector>using namespace std;const int INF=0x3f3f3f3f;const int MAXN=30010;struct qnode{    int v;    int c;    qnode(int _v=0,int _c=0):v(_v),c(_c){}    bool operator <(const qnode &r)const    {        return c>r.c;    }};struct Edge{    int v,cost;    int next;};Edge edge[150005];int tol;int head[MAXN];bool vis[MAXN];int dist[MAXN];int n,m;void Dijkstra(){    memset(vis,false,sizeof(vis));    for(int i=1;i<=n;i++)dist[i]=INF;    priority_queue<qnode>que;    while(!que.empty())que.pop();    dist[1]=0;    que.push(qnode(1,0));    qnode tmp;    while(!que.empty())    {        tmp=que.top();        que.pop();        int u=tmp.v;        if(vis[u])continue;        vis[u]=true;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v;            int cost=edge[i].cost;            if(!vis[v]&&dist[v]>dist[u]+cost)            {                dist[v]=dist[u]+cost;                que.push(qnode(v,dist[v]));            }        }    }    printf("%d\n",dist[n]);}int main(){    while(scanf("%d%d",&n,&m)==2)    {        tol=0;        memset(head,-1,sizeof(head));        int A,B,C;        for(int i = 0;i < m;i++)        {            scanf("%d%d%d",&A,&B,&C);            edge[i].next = head[A];            head[A] = i;            edge[i].v = B;            edge[i].cost = C;        }        Dijkstra();    }    return 0;}