poj 3169 Layout 【差分约束】



Layout

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8249 Accepted: 3949

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 11 3 102 4 202 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.

题意：有n头排成一条线的的牛 和 ML + MD个限制条件，ML个条件表示 a , b两头牛之间距离最多为d，MD个条件表示a, b头牛之间距离最少为d。 让你求出第一头到第n头牛的最大距离，若不存在最大距离（存在负环）输出-1，若距离可以任意输出-2，否则输出最大距离。 题目允许一个位置有多头牛。

明显的差分约束

思路：首先对于任一两头相邻的牛，它们之间距离大于或等于0； 其次，由ML个条件得到 S[b] - S[a] <= d ,最后由MD个条件得到S[a] - S[b] <= -d。

根据这三个条件建图 并让1为源点求到n的最短路即可。若存在负环跳出并输出-1，若最后的最短路为无穷大(初始化都是无穷大)输出-2，反之输出最短路。

AC代码：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#define LL long long#define MAX 1000+10#define INF 10000000+10using namespace std;struct Edge {int from, to, val, next;}edge[1000000+10];int n, big, least;int head[MAX], top;int dist[MAX];int used[MAX];//入队次数bool vis[MAX];//标记是否入队 void init(){top = 0;memset(head, -1, sizeof(head)); } void addEdge(int u, int v, int w){Edge E = {u, v, w, head[u]};edge[top] = E;head[u] = top++;}void getMap(){int a, b, c;for(int i = 1; i < n; i++)addEdge(i+1, i, 0);while(big--){scanf("%d%d%d", &a, &b, &c);addEdge(a, b, c);} while(least--){scanf("%d%d%d", &a, &b, &c);addEdge(b, a, -c);} } void SPFA(){queue<int> Q;for(int i = 1; i <= n; i++){dist[i] = i==1 ? 0 : INF;vis[i] = false;used[i] = 0;}used[1] = 1;vis[1] = true;Q.push(1);while(!Q.empty()){int u = Q.front();Q.pop();vis[u] = false;for(int i = head[u]; i != -1; i = edge[i].next){Edge E = edge[i];if(dist[E.to] > dist[u] + E.val){dist[E.to] = dist[u] + E.val;if(!vis[E.to]){vis[E.to] = true;used[E.to]++;if(used[E.to] > n)//存在负环 {printf("-1\n");return ;} Q.push(E.to);} }}}if(dist[n] == INF) printf("-2\n");//无穷远 elseprintf("%d\n", dist[n]);} int main(){while(scanf("%d%d%d", &n, &big, &least) != EOF){init();getMap();SPFA();}return 0;}