Candies(差分约束_栈+SPFA)

CandiesCrawling in process...Crawling failedTime Limit:1500MS    Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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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 integersN 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 throughN. snoopy and flymouse were always numbered 1 and N. Then followM lines each holding three integers A, B and c in order, meaning that kidA 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 21 2 52 1 4

Sample Output

5

Hint

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

被坑哭了的一道题。上几天学的SPFA+前向星，这题果断用上，数据都对，然后高兴的交上去，TLE，然后吓哭了。后来又用了slf优化一下，还是TLE，这真的是哭死了。然后看了一下poj的讨论区，大牛们说要用栈，sad  又去看的栈，幸亏和队列的用法差不多。

题意：给n个人派糖果，给出m组数据，每组数据包含A，B，c  三个数，A的糖果数比B少的个数不多于c,即B的糖果数 - A的糖果数<= c 。最后求n 比 1 最多多多少糖果。

思路：这是一题典型的差分约束题。不妨将糖果数当作距离，把相差的最大糖果数看成有向边AB的权值，我们得到 dis[B]-dis[A]<=w(A,B)。看到这里，我们联想到求最短路时的松弛技术，即if(dis[B]>dis[A]+w(A,B), dis[B]=dis[A]+w(A,B)。即是满足题中的条件dis[B]-dis[A]<=w(A,B），由于要使dis[B] 最大，所以这题可以转化为最短路来求。（语言组织有点差，只好用了一下别的大牛的思路）

ps：博客里有一篇转载的最短路的，看不明白的可以去学学

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <queue>#include <stack>#include <algorithm>#define inf 9999999using namespace std;int head[30010];int dis[30010];int vis[30010];int n,m,i;int cnt;struct node{    int v,w;    int next;} edge[150000];void add(int u,int v,int w)//加边{    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;}void SPFA(){    memset(dis,inf,sizeof(dis));    memset(vis,0,sizeof(vis));    stack<int>G;//队列    dis[1]=0;    vis[1]=1;    G.push(1);    while(!G.empty())    {        int u=G.top();        G.pop();        vis[u]=0;        for(i=head[u]; i!=-1; i=edge[i].next)        {            int v=edge[i].v;            if(dis[v]>dis[u]+edge[i].w)            {                dis[v]=dis[u]+edge[i].w;                if(!vis[v])                {                    vis[v]=1;                    G.push(v);                }            }        }    }    printf("%d\n",dis[n]);}int main(){    int u,v,w;    while(~scanf("%d %d",&n,&m))    {        cnt=0;        memset(head,-1,sizeof(head));        for(i=1; i<=m; i++)        {            scanf("%d %d %d",&u,&v,&w);            add(u,v,w);        }        SPFA();    }    return 0;}