| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
再写一次,还是不太理解这题常规条件的处理问题,能够帮忙解释下吗?其实如何构图来解决这个问题和为什么用bellmanford解决这个问题我都十分明白了,就是不明白下面列出的一个问题,希望知道的帮忙解答下啊,谢谢!
#include<stdio.h>
#include<algorithm>
using namespace std;
#define MAX 60000
#define MAXINT 1000000
struct EDGE
{
int st,ed,val;
}edge[MAX];
int dis[MAX],n,min,max;
int bellman_ford()
{
int i,k;
for(i=min;i<=max;i++)
dis[i]=-MAXINT;
dis[min]=0;
bool over;
for(k=0;k<=max-min;k++)
{
over=true;
for(i=0;i<n;i++)
if(dis[edge[i].st]!=-MAXINT&&dis[edge[i].st]+edge[i].val>dis[edge[i].ed])
{
dis[edge[i].ed]=dis[edge[i].st]+edge[i].val;
over=false;
}
for(i=min;i<max;i++)
if(dis[i]!=-MAXINT&&dis[i]>dis[i+1])
{
dis[i+1]=dis[i];
over=false;
}
for(i=max;i>min;i--)
if(dis[i]!=-MAXINT&&dis[i]-1>dis[i-1])
{
dis[i-1]=dis[i]-1;
over=false;
}
if(over)
break;
}
return dis[max];
}
int main()
{
int i;
while(scanf("%d",&n)!=EOF)
{
min=MAXINT;
max=0;
for(i=max;i<n;i++)
{
scanf("%d%d%d",&edge[i].st,&edge[i].ed,&edge[i].val);
edge[i].ed++;
if(edge[i].ed>max)
max=edge[i].ed;
if(edge[i].st<min)
min=edge[i].st;
}
printf("%d\n",bellman_ford());
}
return 0;
}
这是晚上的代码,其中下面这段代码是为了解决
对于自然数x,有0 < = F(x+1) - F(x) <= 1,即F(x+1) - F(x) <= 1 && F(x) - F(x+1) <= 0 这个常规约束条件的。不明白为什么要一个升序,一个降序来更新数据,大家都升序来更新不可以吗?希望能指导一下我啊,谢谢!
for(i=min;i<max;i++)
if(dis[i]!=-MAXINT&&dis[i]>dis[i+1])
{
dis[i+1]=dis[i];
over=false;
}
for(i=max;i>min;i--)
if(dis[i]!=-MAXINT&&dis[i]-1>dis[i-1])
{
dis[i-1]=dis[i]-1;
over=false;
}
Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator