POJ 3169 Layout （图论-差分约束）

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6574		Accepted: 3177

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

field=source&key=USACO+2005+December+Gold" style="text-decoration:none">USACO 2005 December Gold

题目大意：

n头奶牛按1到n排好序，md个限制及mt个限制,md行表示奶牛A与奶牛B相差最多D，mt个限制奶牛A与奶牛B相差最少D，问你奶牛1与奶牛n最多相差多少？

解题思路：

限制条件 ：

1、相邻奶牛之间，编号大的距离大于编号小的，即 dist[1]-dist[2]<=0。dist[2]-dist[3]<=0，dist[3]-dist[4]<=0。。。。dist[n-1]-dist[n]<=0

2、md个限制 A与奶牛B相差最多D，dist[B]-dist[A]<=D

3、mt个限制奶牛A与奶牛B相差最少D，dist[B]-dist[A]>=D 。即 dist[B]-dist[A]<=D

v-u<=c，即加入 u->v=c 的单向边

有了这些元素。就能够用差分约束来解了。哈哈，是不是非常easy。

差分约束学习能够參考：http://www.cnblogs.com/void/archive/2011/08/26/2153928.html

解题代码：

#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;

const int maxn=1100;
const int maxm=41000;
const int inf=0x3f3f3f3f;

struct edge{
    int u,v,w,next;
}e[maxm];

int head[maxn],dist[maxn],cnt;
int n;

void initial(){
    cnt=0;
    for(int i=0;i<=n;i++) head[i]=-1;
}

void adde(int u,int v,int w){
    e[cnt].u=u,e[cnt].v=v,e[cnt].w=w,e[cnt].next=head[u],head[u]=cnt++;
}

void input(){
    int m,t;
    scanf("%d%d",&m,&t);
    while(m-- >0){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        adde(u,v,w);
    }
    while(t-- >0){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        adde(v,u,-w);
    }
    for(int i=1;i<=n;i++){
        if(i>=2) adde(i,i-1,0);
        adde(0,i,0);
    }
}

bool spfa(int from){
    int s=from,num[maxn];
    bool visited[maxn];
    for(int i=0;i<=n;i++){
        num[i]=0;
        dist[i]=inf;
        visited[i]=false;
    }
    queue <int> q;
    q.push(s);
    visited[s]=true;
    dist[s]=0;
    while(!q.empty()){
        s=q.front();
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            int d=e[i].v;
            if(dist[d]>dist[s]+e[i].w){
                dist[d]=dist[s]+e[i].w;
                if(!visited[d]){
                    visited[d]=true;
                    q.push(d);
                    num[d]++;
                    if(num[d]>n) return false;
                }
            }
        }
        visited[s]=false;
    }
    return true;
}

void solve(){
    if(spfa(0)){
        if(spfa(1)){
            if(dist[n]==inf) printf("-2\n");
            else printf("%d\n",dist[n]);
        }
    }else printf("-1\n");
}

int main(){
    while(scanf("%d",&n)!=EOF){
        initial();
        input();
        solve();
    }
    return 0;
}