poj 3169 Layout

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6662		Accepted: 3220

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

差分約束系統的典型，可以利用bellman-ford或者spfa解決，spfa效率更高，在判負環的時候開一個ord數組，判斷位置是否超過邊數：

判斷負權迴路的方案很多

1、記錄每個結點進隊次數，超過|V|次表示有負環；

2、記錄這個結點在路徑中處於的位置ord[i]，每次更新的時候ord[i]=ord[x]+1，若超過|V|則表示有負環

看錯頂點個數，內存超出限制，~~~~(>_<)~~~~ ，下邊是AC代碼：

/*
13074017	motefly	3169	Accepted	4644K	110MS	G++	1607B	2014-07-14 11:17:08
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAX_V=1005;
const int INF=0x3f3f3f3f;
int N,ML,MD;
int cnt;
int G[MAX_V][MAX_V];
int d[MAX_V],ord[MAX_V],mark[MAX_V];
void init()
{
    memset(G,INF,sizeof(G));
    scanf("%d%d%d",&N,&ML,&MD);
    for(int i=1;i<=N;i++)
        G[i][i-1]=0;
    for(int i=0;i<ML;i++)
    {
        int f,t,c;
        scanf("%d%d%d",&f,&t,&c);
        G[f][t]=c;
        if(f+1!=t&&t+1!=f)
            cnt++;
    }
    for(int i=0;i<MD;i++)
    {
        int f,t,c;
        scanf("%d%d%d",&f,&t,&c);
        G[t][f]=-c;
        if(f+1!=t&&t+1!=f)
            cnt++;
    }
    cnt+=N-1;
}


int spfa(int v)
{
    for(int i=1;i<=N;i++)
        d[i]=INF;
    queue<int> q;
    q.push(v);
    ord[v]=1;
    mark[v]=1;
    d[v]=0;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        mark[u]=0;
        for(int i=1;i<=N;i++)
        {
            if(G[u][i]!=INF)
            {
                if(d[u]+G[u][i]<d[i])
                {
                    d[i]=d[u]+G[u][i];
                    ord[i]=ord[u]+1;
                    if(!mark[i])
                    {
                        q.push(i);
                        mark[i]=1;
                    }
                    if(ord[i]>cnt)
                        return -1;
                }
            }
        }
    }
    if(d[N]==INF)
        return -2;
    return d[N];
}

int main()
{
    init();
    cout<<spfa(1)<<endl;
    return 0;
}