POJ 3169 Layout （Bellman-Ford、差分约束）

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7834		Accepted: 3753

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

求牛1-N 其实最最大距离就是1-N的满足条件的最短路

由最短路性质可知 对于所有的边(u, v)  均有d[u] + cost[u][v] >= d[v]

那么由题目已知  必须顺序排序 但是可以多牛共存一点

即 d[i+1] >= d[i] → d[i+1] + 0 >= d[i] (从i+1连一条权为0的边到i)

喜欢的牛 B - A <= D → A + D >= B (从A连一条权为D的边到B)

讨厌的牛 B - A >= D → B - D >= A (从B连一条权为-D的边到A)

由于存在负权 我们可以用Bellman-Ford来算最短路 O(EV)的复杂度题目规模完全够

AC代码如下：

//
//  POJ 3169 Layout
//
//  Created by TaoSama on 2015-03-20
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x,y) memset(x, y, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n, l, d, dp[1005];
struct Edge {
	int to, cost;
};
vector<Edge> G[21005];

int BellmanFord() {
	memset(dp, 0x3f, sizeof dp); dp[1] = 0;
	for(int k = 1; k <= n; ++k) {
		for(int i = 1; i <= n; ++i) {
			for(int j = 0; j < G[i].size(); ++j) {
				Edge &e = G[i][j];
				dp[e.to] = min(dp[e.to], dp[i] + e.cost);
				//printf("dp[%d]: %d\n",e.to, dp[e.to]);
			}
		}
	}
	return dp[n];
}

int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	ios_base::sync_with_stdio(0);

	scanf("%d%d%d", &n, &l, &d);
	int a, b, c;
	for(int i = 1; i < n; ++i)
		G[i + 1].push_back((Edge) {i, 0});
	for(int i = 1; i <= l; ++i) {
		scanf("%d%d%d", &a, &b, &c);
		G[a].push_back((Edge) {b, c});
	}
	for(int i = 1; i <= d; ++i) {
		scanf("%d%d%d", &a, &b, &c);
		G[b].push_back((Edge) {a, -c});
	}
	int ans = BellmanFord();
	if(dp[1] < 0) ans = -1;
	else if(ans == INF)  ans = -2;
	printf("%d\n", ans);
	return 0;
}