poj 3169 Layout（差分约束）

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6549		Accepted: 3168

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

Sample Output

解题思路：

1、构造差分约束系统：

A。B距离不超过D则B-A<=D，

A,B距离至少为D则A-B<=-D.

2、若有解则求1和N之间的最短路径：

比如：A-B<=D1 , B-C<=D2, A-C<=D3 不等式相加得：A-C<=min(D3,D1+D2)。而当A-B<=D时。我们建的边是B->A的。所以我们仅仅要求出1到N的最短距离就可以，假设没有最短距离。则输出-2.

#include <iostream>

#include <cstdio>

#include <queue>

using namespace std;

const int maxne = 1000000;

const int maxnn = 1010;

const int INF = 0x3f3f3f3f;

struct edge{

    int u , v , d;

    edge(int a = 0 , int b = 0 , int c = 0){

        u = a , v = b , d = c;

    }

}e[maxne];

int head[maxnn] , next[maxne] , cnt , dis[maxnn] , vis[maxnn] , vt[maxnn];

int N , ML , MD;

void add(int u , int v , int d){

    e[cnt] = edge(u , v , d);

    next[cnt] = head[u];

    head[u] = cnt++;

}

void initial(){

    for(int i = 0; i < maxnn; i++) head[i] = -1 , dis[i] = INF , vis[i] = 0 , vt[i] = 0;

    for(int i = 0; i < maxne; i++) next[i] = -1;

    cnt = 0;

    for(int i = 1; i < N; i++){

        add(0 , i , 0);

        add(i+1 , i , 0);

    }

    add(0 , N , 0);

    dis[0] = 0;

}

void readcase(){

    int u , v , d;

    while(ML--){

        scanf("%d%d%d" , &u , &v , &d);

        add(u , v , d);

    }

    while(MD--){

        scanf("%d%d%d" , &u , &v , &d);

        add(v , u , -1*d);

    }

}

bool SPFA(int start){

    queue<int> q;

    q.push(start);

    vt[start]++;

    while(!q.empty()){

        int u = q.front();

        q.pop();

        vis[u] = 0;

        int n = head[u];

        if(vt[u] > N+1){

            return false;

        }

        while(n != -1){

            int v = e[n].v;

            if(dis[v] > dis[u]+e[n].d){

                dis[v] = dis[u]+e[n].d;

                if(vis[v] == 0){

                    vt[v]++;

                    q.push(v);

                    vis[v] = 1;

                }

            }

            n = next[n];

        }

    }

    return true;

}

void computing(){

    if(!SPFA(0)){

        printf("-1\n");

    }else{

        for(int i = 0; i < maxnn; i++) dis[i] = INF;

        dis[1] = 0;

        SPFA(1);

        if(dis[N] == INF){

            printf("-2\n");

        }else{

            printf("%d\n" , dis[N]);

        }

    }

}

int main(){

    while(scanf("%d%d%d" , &N , &ML , &MD) != EOF){

        initial();

        readcase();

        computing();

    }

    return 0;

}

巴特西

poj 3169 Layout（差分约束）

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