POJ-3169 Layout

Layout

Time Limit: 1000MS		Memory Limit: 65536K

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.

————————————————————水水的分割线————————————————————

思路：求最大值，跑最短路。根据小于等于关系建图，注意序号大的牛在序号小的牛的后面。（S[i] - S[i-1] >= 0）

然后以牛1为源点，dis数组的含义就是和牛1之间的距离。如果为INF，说明距离可以任意大。如果有负环，说明不存在解。

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <climits>
#include <iostream>
#define INF 0x3f3f3f3f
using namespace std;
/****************************************/
const int N = 1005, M = 1e5;
struct Node {
	int v, w, next;
}edge[M];
int n, tot, q[N], head[N], dis[N], enq[N];
bool inq[N];

void add(int u, int v, int w)
{
	edge[tot].v = v;
	edge[tot].w = w;
	edge[tot].next = head[u];
	head[u] = tot++;
}

bool spfa(int st)
{
	for(int i = 1; i <= n; i++) {
		dis[i] = INF;
		inq[i] = enq[i] = 0;
	}
	dis[st] = 0;
	int fron = 0, rear = 0;
	q[rear++] = st;
	inq[st] = 1; enq[st]++;
	while(fron < rear) {
		int u = q[fron%N]; fron++;
		inq[u] = false;
		for(int i = head[u]; i != -1; i = edge[i].next) {
			int v = edge[i].v;
			if(dis[v] > dis[u] + edge[i].w) {
				dis[v] = dis[u] + edge[i].w;
				if(!inq[v]) {
					q[rear%N] = v; rear++;
					inq[v] = true;
					enq[v]++;
					if(enq[v] > n) return false;	
				}
			}
		}
	}
	return true;
}

int main()
{
#ifdef J_Sure
//	freopen("000.in", "r", stdin);
//	freopen(".out", "w", stdout);
#endif
	int L, D;
	scanf("%d%d%d", &n, &L, &D);
	memset(head, -1, sizeof(head));
	tot = 0;
	int a, b, d;
	while(L--) {
		scanf("%d%d%d", &a, &b, &d);
		//b-a <= d
		add(a, b, d);
	}
	while(D--) {
		scanf("%d%d%d", &a, &b, &d);
		//b-a >= d
		add(b, a, -d);
	}
	for(int i = 2; i <= n; i++) {
		//s[i] - s[i-1] >= 0
		add(i, i-1, 0);
	}
	bool flag = spfa(1);
	if(dis[n] == INF) printf("-2\n");
	else if(flag) printf("%d\n", dis[n]);
	else printf("-1\n");
	return 0;
}