zoukankan      html  css  js  c++  java
  • (简单) POJ 3169 Layout,差分约束+SPFA。

      Description

      Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).

      Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated.

      Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.
     
      题意就是 Xi-Xj<c 然后求 Xn-X1 的最大值,差分约束问题。。。
      建图然后SPFA 就好。。。
     
    代码如下:
    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <time.h>
        
    using namespace std;
    
    const int MaxN=1010;
    const int MaxM=30100;
    const int INF=1000000009;
    
    struct Edge
    {
        int to,next,cost;
    };
    
    Edge E[MaxM];
    int head[MaxN],Ecou;
    
    bool vis[MaxN];
    int couNode[MaxN];
    
    void init(int N)
    {
        Ecou=0;
        for(int i=1;i<=N;++i)
        {
            head[i]=-1;
            couNode[i]=0;
            vis[i]=0;
        }
    }
    
    void addEdge(int u,int v,int w)
    {
        E[Ecou].to=v;
        E[Ecou].cost=w;
        E[Ecou].next=head[u];
        head[u]=Ecou++;
    }
    
    bool SPFA(int lowcost[],int N,int start)
    {
        int t,v;
        queue <int> que;
    
        for(int i=1;i<=N;++i)
            lowcost[i]=INF;
        lowcost[start]=0;
    
        que.push(start);
        couNode[start]=1;
        vis[start]=1;
    
        while(!que.empty())
        {
            t=que.front();
            que.pop();
    
            vis[t]=0;
    
            for(int i=head[t];i!=-1;i=E[i].next)
            {
                v=E[i].to;
                
                if(lowcost[v]>lowcost[t]+E[i].cost)
                {
                    lowcost[v]=lowcost[t]+E[i].cost;
    
                    if(!vis[v])
                    {
                        vis[v]=1;
                        couNode[v]+=1;
                        que.push(v);
    
                        if(couNode[v]>N)
                            return 0;
                    }
                }
            }
        }
    
        return 1;
    }
    
    int ans[MaxN];
    
    int main()
    {
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
        
        int N,ML,MD;
        int a,b,c;
    
        scanf("%d %d %d",&N,&ML,&MD);
    
        init(N);
    
        for(int i=1;i<=ML;++i)
        {
            scanf("%d %d %d",&a,&b,&c);
    
            addEdge(a,b,c);
        }
    
        for(int i=1;i<=MD;++i)
        {
            scanf("%d %d %d",&a,&b,&c);
    
            addEdge(b,a,-c);
        }
    
        for(int i=1;i<=N-1;++i)
            addEdge(i+1,i,0);
    
        if(!SPFA(ans,N,1))
            printf("-1
    ");
        else if(ans[N]!=INF)
            printf("%d
    ",ans[N]);
        else
            printf("-2
    ");
    
        return 0;
    }
    View Code
  • 相关阅读:
    大话设计模式之代理模式
    大话设计模式之装饰者模式
    策略模式与简单工厂模式
    一个简单的使用Quartz和Oozie调度作业给大数据计算平台执行
    oozie JAVA Client 编程提交作业
    HashMap分析及散列的冲突处理
    cmp排序hdoj 1106排序
    定义member【C++】cstddef中4个定义
    目录启动CXF启动报告LinkageError异常以及Java的endorsed机制
    算法代码[置顶] 机器学习实战之KNN算法详解
  • 原文地址:https://www.cnblogs.com/whywhy/p/4337864.html
Copyright © 2011-2022 走看看