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Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11612		Accepted: 5550

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. 題目鏈接：http://poj.org/problem?id=3169 題意：有n頭牛按編號順序站一排，即每頭牛都有一個一維坐標，可以相同。現在有一些牛之間有關系，關系好的a，b必須距離小於等於dl；關系不好的a，b必須距離大於等於dd。求牛1和牛n的最大距離。 思路：最短路問題：<u,v> d[u]+d>=d[v]。 n頭牛按編號順序站一排，則d[i+1]>=d[i]，即編號大的牛的坐標大於等於編號小的牛。關系好的牛a，牛b，則d[a]+d>=d[b];關系不好的牛a，牛b，則d[a]+d<=d[b]，即d[b]+(-d)>=d[a]。求約束下的最大距離。最短路也可以理解為約束下的最大解。 因為存在負權值，所以有可能存在負權值回路，所以dijkstra算法不能使用，直接使用ford算法。存在負權值回路輸出-1，d[n]=inf輸出-2，其他情況直接輸出d[n]。 代碼：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
const int maxn=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e13+7;
struct edge
{
    int from,to;
    int cost;
};
int cou=0;
edge es[maxn];
vector<edge>G[maxn];
int used[maxn];
priority_queue<P,vector<P>,greater<P> >que;
void addedge(int u,int v,int w)
{
    cou++;
    edge e;
    e.from=u,e.to=v,e.cost=w;
    es[cou].from=u,es[cou].to=v,es[cou].cost=w;
    G[u].push_back(e);
}
int n,ml,md;
int al[maxn],bl[maxn],dl[maxn];
int ad[maxn],bd[maxn],dd[maxn];
int d[maxn];
void ford()
{
    for(int i=1; i<=n; i++) d[i]=inf;
    d[1]=0;
    for(int t=1; t<n; t++)
    {
        for(int i=1; i<n; i++)
            if(d[i+1]<inf) d[i]=min(d[i],d[i+1]);
        for(int i=1; i<=ml; i++)
            if(d[al[i]]<inf) d[bl[i]]=min(d[bl[i]],d[al[i]]+dl[i]);
        for(int i=1; i<=md; i++)
            if(d[bd[i]]<inf)  d[ad[i]]=min(d[ad[i]],d[bd[i]]-dd[i]);
    }
    if(d[1]<0) cout<<-1<<endl;
    else if(d[n]>=inf) cout<<-2<<endl;
    else cout<<d[n]<<endl;
}
int main()
{
    int a,b,d;
    scanf("%d%d%d",&n,&ml,&md);
    for(int i=1; i<n; i++) addedge(i+1,i,0);
    for(int i=1; i<=ml; i++)
        scanf("%d%d%d",&al[i],&bl[i],&dl[i]);
    for(int i=1; i<=md; i++)
        scanf("%d%d%d",&ad[i],&bd[i],&dd[i]);
    ford();
    return 0;
}
/*
4 3 0
1 3 10
2 4 20
2 3 3
*/

最短路

POJ 3169.Layout 最短路