POJ 3169-Layout（差分約束系統-入門裸題）

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.

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.

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.

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

27

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.

USACO 2005 December Gold

有編號為1~N的N頭母牛，多個母牛的座標位置可能相同。

有ML對母牛相互喜歡，A對B的距離不能大於D；

有MD對母牛相互嫌棄，A對B的距離不能小於D；

求第1頭與第N頭母牛之間的最大距離；如果不存在，輸出“-1”；如果距離無約束，輸出“-2”。

紅果果的差分約束系統的裸題。

Db-Da≤d，則a~b權值為d；

Db-Da≥d，轉化成Da-Db≤-d，則b~a權值為-d。

建一個有向圖，喜歡那麼距離為正，嫌棄的話距離為負。

直接SPFA求解dist[n]即1~N奶牛的距離。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <iomanip>
#include <algorithm>
#define MAXN 10010
#define INF 0xfffffff
using namespace std;

struct ArcNode
{
    int to;
    int weight;
    ArcNode *next;
};
queue<int> Q;//佇列中的節點為頂點序號
int n;//頂點個數
ArcNode * List[MAXN];//每個頂點的邊連結串列表頭指標
int inq[MAXN];//每個頂點是否在佇列中的標誌
int dist[MAXN],path[MAXN];
int cnt[MAXN];
bool SPFA(int src)
{
    memset(cnt,0,sizeof(cnt));
    memset(inq,0,sizeof(inq));
    int i,u;//u為佇列頭頂點序號
    ArcNode * temp;
    for(i=1; i<=n; ++i)//初始化
    {
        dist[i]=INF;
        path[i]=src;
        inq[i]=0;
    }
    dist[src]=0;
    path[src]=src;
    ++inq[src];
    Q.push(src);
    ++cnt[src];
    while(!Q.empty())
    {
        u=Q.front();
        Q.pop();
        --inq[u];
        if(cnt[u]>n) return true;//存在負權迴路
        temp=List[u];
        while(temp!=NULL)
        {
            int v=temp->to;
            if(dist[v]>dist[u]+temp->weight)
            {
                dist[v]=dist[u]+temp->weight;
                if(!inq[v])
                {
                    Q.push(v);
                    ++inq[v];
                    ++cnt[v];
                }
            }
            temp=temp->next;
        }
    }
    return false;
}
int main()
{
    int a,b;
    int u,v,w;
    cin>>n>>a>>b;
    memset(List,0,sizeof(List));
    ArcNode *temp;
    while(!Q.empty()) Q.pop();
    while(a--)
    {
        cin>>u>>v>>w;
        temp=new ArcNode;
        temp->to=v;//構造鄰接表
        temp->weight=w;
        temp->next=NULL;
        if(List[u]==NULL) List[u]=temp;
        else
        {
            temp->next=List[u];
            List[u]=temp;
        }
    }
    while(b--)//負權路
    {
        cin>>u>>v>>w;
        temp=new ArcNode;
        temp->to=u;//構造鄰接表
        temp->weight=-w;
        temp->next=NULL;
        if(List[v]==NULL) List[v]=temp;
        else
        {
            temp->next=List[v];
            List[v]=temp;
        }
    }
    bool flag=SPFA(1);//求源點到其他頂點的最短路徑
    /*for(i=0; i<=n; ++i)
        cout<<dist[i]<<" ";
    cout<<endl;*/
    if(flag) puts("-1");//存在負環
    else
    {
        if(dist[n]==INF) puts("-2");//1~N未更新說明木有約束
        else cout<<dist[n]<<endl;
    }
    return 0;
}
/*
4 2 1
1 3 10
2 4 20
2 3 3
*/