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POJ3169 Layout【SPFA】【差分约束】

2024-11-19 15:07:39   203人阅读

Layout
Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 7608Accepted: 3653
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.


Input

Line 1: Three space-separated integers: N, ML, and MD. 

Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart. 

Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.


Output

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


Sample Input

4 2 1
1 3 10
2 4 20
2 3 3


Sample Output

27


Hint

Explanation of the sample: 
There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart. 

The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.


Source

USACO 2005 December Gold


题目大意：奶牛喜欢站成一排吃饲料，有的奶牛喜欢靠在一起，他们最多距离D米远。

有的奶牛不喜欢靠在一起，他们最少距离D米远才可以。给你一系列的约束条件，问：

如果在上述条件下，奶牛站不成一排，则输出"-1"，如果能站成一排，但是第1头牛~

第N头牛的距离无限远，则输出"-2"，如果满足条件，并且第1头牛~第N头牛之间存在

实际距离，则输出第1头牛~到第N头牛之间的能达到的最远距离。

思路：直接的差分约束系统，设牛u和牛v的位置为u和v。

第1条：设牛u和牛v最多距离w米远，转换为：v - u <= w。

第2条：设牛u和牛v最少距离w米远，转换为：u - v <= -w。

还有一个隐含条件：两头牛之间的距离最少为0米，转换为 i - (i+1) <= 0。遍历加入。

后来证明不加也可以A过。。。因为这个条件(-1 <= 0)本来就成立嘛

然后SPFA判断并求距离


#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#define INF 0x7fffffff
using namespace std;
const int MAXN = 1100;
const int MAXM = 30030;

struct EdgeNode
{
    int to;
    int w;
    int next;
}Edges[MAXM];

int Head[MAXN],Dist[MAXN],vis[MAXN],outque[MAXN],id;

void AddEdges(int u,int v,int w)
{
    Edges[id].to = v;
    Edges[id].w = w;
    Edges[id].next = Head[u];
    Head[u] = id++;
}
void SPFA(int s,int N)
{
    int ans = 0;
    memset(vis,0,sizeof(vis));
    memset(outque,0,sizeof(outque));
    for(int i = 1; i <= N; ++i)
        Dist[i] = INF;
    Dist[s] = 0;
    vis[s] = 1;
    queue<int> Q;
    Q.push(s);
    while( !Q.empty() )
    {
        int u = Q.front();
        Q.pop();
        vis[u] = 0;
        outque[u]++;
        if(outque[u] > N+1)
        {
            ans = -1;
            break;
        }
        for(int i = Head[u]; i != -1; i = Edges[i].next)
        {
            int temp = Dist[u] + Edges[i].w;
            if(temp < Dist[Edges[i].to])
            {
                Dist[Edges[i].to] = temp;
                if( !vis[Edges[i].to])
                {
                    vis[Edges[i].to] = 1;
                    Q.push(Edges[i].to);
                }
            }
        }
    }

    if(ans == -1)
        printf("-1\n");
    else if(Dist[N] == INF)
        printf("-2\n");
    else
        printf("%d\n",Dist[N]-Dist[1]);
}

int main()
{
    int N,ML,MD,u,v,w;
    while(~scanf("%d%d%d", &N, &ML, &MD))
    {
        memset(Head,-1,sizeof(Head));
        id = 0;
        for(int i = 0; i < ML; ++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            AddEdges(u,v,w);
        }
        for(int i = 0; i < MD; ++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            AddEdges(v,u,-w);
        }
//这里不加也可以
//        for(int i = 1; i < N; ++i)
//            AddEdges(i+1,i,0);    
        SPFA(1,N);
    }

    return 0;
}


POJ3169 Layout【SPFA】【差分约束】


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