HDU 3592 World Exhibition（线性差分约束，spfa跑最短路+判断负环）-优快云博客

本文探讨了一种有趣的排队问题，源自世博会参观者的需求。问题要求在满足特定的人际距离约束下，计算队列中首尾两人之间的最大可能距离。通过使用SPFA算法和图论中的负环检测，文章提供了详细的代码实现和解析。

World Exhibition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2162 Accepted Submission(s): 1063

Problem Description

Nowadays, many people want to go to Shanghai to visit the World Exhibition. So there are always a lot of people who are standing along a straight line waiting for entering. Assume that there are N (2 <= N <= 1,000) people numbered 1..N who are standing in the same order as they are numbered. It is possible that two or more person line up at exactly the same location in the condition that those visit it in a group.

There is something interesting. Some 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 X (1 <= X <= 10,000) constraints describes which person like each other and the maximum distance by which they may be separated; a subsequent list of Y constraints (1 <= Y <= 10,000) tells which person 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 person 1 and person N that satisfies the distance constraints.

Input

First line: An integer T represents the case of test.

The next line: Three space-separated integers: N, X, and Y.

The next X lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= N. Person A and B must be at most C (1 <= C <= 1,000,000) apart.

The next Y lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= C. Person A and B must be at least C (1 <= C <= 1,000,000) apart.

Output

For each line: A single integer. If no line-up is possible, output -1. If person 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between person 1 and N.

Sample Input

1 4 2 1 1 3 8 2 4 15 2 3 4

Sample Output

Author

alpc20

Source

2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

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分析：

负环的定义：

有向图中存在一个环，其权值和为负数

这题跟poj3169一模一样

解析：

https://www.cnblogs.com/yinbiao/p/10001772.html

code：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<math.h>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
#define INF 9999999999
#define me(a,x) memset(a,x,sizeof(a))
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};

int getval()
{
    int ret(0);
    char c;
    while((c=getchar())==' '||c=='\n'||c=='\r');
    ret=c-'0';
    while((c=getchar())!=' '&&c!='\n'&&c!='\r')
        ret=ret*10+c-'0';
    return ret;
}
void out(int a)
{
    if(a>9)
        out(a/10);
    putchar(a%10+'0');
}

#define max_v 1005
struct node
{
    int v;
    LL w;
    node(int vv=0,LL ww=0):v(vv),w(ww) {}
};
LL dis[max_v];
int vis[max_v];
int cnt[max_v];
vector<node> G[max_v];
queue<int> q;

void init()
{
    for(int i=0; i<max_v; i++)
    {
        G[i].clear();
        dis[i]=INF;
        vis[i]=0;
        cnt[i]=0;
    }
    while(!q.empty())
        q.pop();
}

int spfa(int s,int n)
{
    vis[s]=1;
    dis[s]=0;
    q.push(s);
    cnt[s]++;

    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;

        for(int j=0; j<G[u].size(); j++)
        {
            int v=G[u][j].v;
            LL w=G[u][j].w;

            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(vis[v]==0)
                {
                    q.push(v);
                    cnt[v]++;
                    vis[v]=1;

                    if(cnt[v]>n)
                        return 0;
                }
            }
        }
    }
    return 1;
}
int f(int u,int v)
{
    for(int j=0; j<G[u].size(); j++)
    {
        if(G[u][j].v==v)
            return 0;
    }
    return 1;
}
int main()
{
    int n,a,b;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d",&n,&a,&b);
        init();
        int x,y,w;
        while(a--)
        {
            scanf("%d %d %d",&x,&y,&w);
            if(f(x,y))
                G[x].push_back(node(y,w));
        }
        while(b--)
        {
            scanf("%d %d %d",&x,&y,&w);
            if(f(y,x))
                G[y].push_back(node(x,-w));
        }
        int flag=spfa(1,n);
        if(flag==0)
        {
            printf("-1\n");
        }
        else if(dis[n]<INF)
        {
            printf("%lld\n",dis[n]);
        }
        else
        {
            printf("-2\n");
        }
    }
    return 0;
}