poj 3169 差分约束

Layout

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10068 Accepted: 4840

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：

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.

就是说如果 1  n 没有连接在一起，就输出 -1

如果距离可以无限远就输出 -2

有限距离，计算出最大的距离就好了

无限远用 dis 【n】==INF 就好了

本题队列比栈实现SPFA快一些，不知道为什么，有些题目栈却快一点

本题中隐含的条件：

 i +1 - i >=0

我一直没有将所有的点这样处理，只是处理输入的点

唉！！！算法理解的还不够透彻啊，继续加油吧

#include<queue>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;int n,ml,md;int head[21000],vis[21000];int dis[21000],num[21000];struct node{    int v,w,next;}path[21000];int top;void add(int u,int v,int w){    path[top].v=v;    path[top].w=w;    path[top].next=head[u];    head[u]=top++;}int SPFA(){    memset(vis,0,sizeof(vis));    memset(num,0,sizeof(num));    for(int i=1;i<=n;i++)        dis[i]=INF;    queue<int>q;    dis[1]=0;    q.push(1);    num[1]++;    while(q.size())    {        int p=q.front();        q.pop();        vis[p]=0;        for(int i=head[p];i!=-1;i=path[i].next){            int t=path[i].v;            if(dis[t]>dis[p]+path[i].w){                dis[t]=dis[p]+path[i].w;                if(vis[t]==0){                    q.push(t);                    num[t]++;                    vis[t]=1;                    if(num[t]>n)                        return -1;                }            }        }    }    if(dis[n]>=INF)        return -2;    return dis[n];}int main(){    top=0;    int a,b,c;    //freopen("in.txt","r",stdin);    memset(head,-1,sizeof(head));    scanf("%d%d%d",&n,&ml,&md);    while(ml--)    {        scanf("%d%d%d",&a,&b,&c);        add(a,b,c);    }    while(md--)    {        scanf("%d%d%d",&a,&b,&c);        add(b,a,-c);    }    for(int i=1;i<=n;i++)        add(i,i-1,0);    int ans=SPFA();    printf("%d\n",ans);    return 0;}