POJ 3159 Candies【差分约束+SPFA】

Candies

Time Limit: 1500MS

 

Memory Limit: 131072K

Total Submissions: 28026

 

Accepted: 7731

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 throughN. 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.

Source

POJ Monthly--2006.12.31, Sempr

 

题目大意：给n个人分糖果，给出m个约数条件，每组包含三个数a，b，c，表示b的糖果数比a的糖果数不能多c个，即b-a<=c

问n比1最多能够多多少糖果。

题目分析：

明显的差分约束系统问题，能够将问题转化为最短路问题。以下给出转化过程及详解：

假设有这样的三个条件：

我们不妨把1,2,3三个人都看成节点，然后设k1，k2，k3为三条有向边的权值，那么1-2这条边能够取值为【0，k1】因为我们要求最大值，那么我们必然要取1-2这条边的权值为k1,2-3这条边的权值为k2，1-3这条边的权值为k3。

将约束条件转化为有向图：

我们再根据约束条件1和约束条件2相加，得到约束条件：1-3<=k1+k2,我们还有1-3<=k3两者同时成立，那么k1+k2和k3这两个数，最小的就是1比3多的糖果数。

我们再根据有向图的最短路松弛过程有：k1+k2与k3相比较，最小的就是3到1的有向图最短路径，那么不难理解，由差分约束问题很容易就转化到了最短路问题上来。

再因为数据比较大，我们采用SPFA+stack+读入外挂（fast IO）来优化这个问题，最后剩下的就是代码实现问题：

#include<stdio.h>#include<string.h>#include<queue>#include<stack>using namespace std;#define INF 0x3f3f3f3fint head[30010];struct EdgeNode{    int to;    int w;    int next;}e[150010];int vis[30010];int dis[30010];int n,m;int cont;void add(int from,int to,int w){    e[cont].to=to;    e[cont].w=w;    e[cont].next=head[from];    head[from]=cont++;}int Scan(){    int res = 0, ch, flag = 0;    if((ch = getchar()) == '-')             //判断正负        flag = 1;    else if(ch >= '0' && ch <= '9')           //得到完整的数        res = ch - '0';    while((ch = getchar()) >= '0' && ch <= '9' )        res = res * 10 + ch - '0';    return flag ? -res : res;}void SPFA(int start){    memset(vis,0,sizeof(vis));    for(int i=0;i<=n;i++)dis[i]=INF;    stack<int >s;    s.push(start);    vis[start]=1;    dis[start]=0;    while(!s.empty())    {        int u=s.top();        //printf("u:%d\n",u);        s.pop();vis[u]=0;        for(int j=head[u];j!=-1;j=e[j].next)        {            int v=e[j].to;            int w=e[j].w;            //printf("%d %d\n",v,w);            if(dis[v]>dis[u]+w)            {                dis[v]=dis[u]+w;                if(vis[v]==0)                {                    s.push(v);                    vis[v]=1;                }            }        }    }    printf("%d\n",dis[n]);}int main(){    while(~scanf("%d%d",&n,&m))    {        cont=0;        memset(head,-1,sizeof(head));        for(int i=0;i<m;i++)        {            int x,y,w;            x=Scan();            y=Scan();            w=Scan();            add(x,y,w);        }        SPFA(1);    }}