POJ - 1062 Expensive bride price (BFS solution)

expensive bride price

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 53123		Accepted: 16003

Description

The young explorer came to an Indian tribe. There he fell in love with the chief's daughter, so he went to the chief to ask for marriage. The chief asked him to use 10,000 gold coins as a dowry before agreeing to marry his daughter to him. The explorer couldn't come up with so many gold coins, so he asked the chief to lower his requirements. The chief said, "Well, if you can get me the high priest's coat, I can get 8,000 gold coins. If you can get his crystal ball, then it will only cost 5,000 gold coins." The explorer ran to the high priest. There, ask him for a leather jacket or a crystal ball, and the high priest asks him to exchange gold coins, or get him something else, and he can lower the price. The explorers then went elsewhere, and others made similar requests, either directly exchanging gold coins, or finding other things to reduce the price. But explorers don't need to trade multiple things for the same thing, because they won't get a lower price. The explorer needs your help now, let him marry his sweetheart with the least amount of gold coins. In addition, what he wants to tell you is that in this tribe, the concept of hierarchy is very strict. There will be no direct contact of any kind, including transactions, between two persons whose status gap exceeds a certain limit. He's an outsider, so he's exempt from these restrictions. But if he trades with someone of lower status, the higher-status person won't trade with him again, they see it as indirect contact, and vice versa. Therefore, you need to provide him with the best solution after considering all the circumstances.
For convenience, we number all items starting from 1, the chief's promise is also regarded as an item, and the number is always 1. Each item has a corresponding price P, the owner's status level L, and a series of substitutes Ti and the "offer" Vi corresponding to the substitute. If the difference between the two's status level exceeds M, they cannot "indirect transaction". You have to use this data to figure out the minimum amount of gold an explorer needs to marry the chief's daughter.

Input

The first line of input is two integers M, N (1 <= N <= 100), which in turn represent the status level gap limit and the total number of items. Next, the descriptions of the N items are given in order from small to large. The description of each item begins with three non-negative integers P, L, X (X < N), which in turn represent the price of the item, the status level of the owner, and the total number of substitutes. Each of the next X lines includes two integers T and V, representing the replacement number and "preferential price", respectively.

Output

Output the minimum required gold coins.

Sample Input

Sample Output

The meaning of the question: This question is to find a shortest path that meets the meaning of the question. Let the starting point be S and point 1 as the destination. Now there is a way from S to 1, but the road may be very long, so some small roads can also be reached. For point 1, each place has a level, and the maximum level difference of the road you walk cannot exceed m.

Idea: Let’s talk about the construction first, start with the example: S-->1 length 10000, there are two paths, 2-->1 length 8000, 3-->1 length 5000; S-->2 length 1000, and there is 1 path, 4-->2 long 200; I believe everyone can understand how the picture is built when you see it here, and the path is equivalent to a shortcut to exchange items.

Because there is also a grade system in the question, the grade difference between any two points on the road you walk cannot exceed m, so use search to record the maximum grade and the minimum grade of each state. See the code explanation for details.

PS: 11 months ago, I just learned the shortest way. When I did this question, I felt it was too difficult to give up. Now when I go back and make up the question, I know how to write it after a while, and I have the ability to do the question. Great improvement, keep going! .

AC code:

#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#define delta 0.98
#define LL long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define ABS(a) a>=0?a:-a
const double eps=1e-7;
const int N=205;
using namespace std;
struct node1
{
    int v,nex,w;
}e[10500];//代码习惯用数组模拟链表 建图
struct node
{
    int s,l,r,num;//s表示当前所在的点，l，r表示这条路上的等级的最小最大值，num表示路长。
};
int head[N],d[N];//d数组存等级
int cnt,S,n,m;//S表示出发点0
void add(int u,int v,int w)
{
    e[cnt].v=v;
    e[cnt].w=w;
    e[cnt].nex=head[u];
    head[u]=cnt++;
}
void init()
{
    mem(head,-1);
    S=cnt=0;
}
int BFS()
{
    int ans=inf;//存最小答案
    queue<node>q;
    node q1,q2;
    q1.l=q1.r=d[1];//，因为终点是1，所以初始等级可以直接设成d[1]；
    q1.num=q1.s=0;
    q.push(q1);
    while(!q.empty())
    {
        q1=q.front();
        q.pop();
        int u=q1.s;
        if(u==1)//如果找到终点
        {
            if(ans>q1.num) ans=q1.num;//跟新答案
            continue;
        }
        for(int i=head[u];~i;i=e[i].nex)
        {
            q2=q1;
            int v=e[i].v;
            if(d[v]<q2.l) q2.l=d[v];//跟新最大最小等级
            if(d[v]>q2.r) q2.r=d[v];
            if(q2.l+m<q2.r) continue;//不符合的剪枝。
            q2.num+=e[i].w;
            if(q2.num>=ans) continue;//路长超过答案的，剪枝
            q2.s=v;
            q.push(q2);
        }
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        init();
        int x,a,b;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&a,&d[i],&x);
            add(S,i,a);//建大路
            while(x--)//建小路
            {
                scanf("%d%d",&a,&b);
                add(a,i,b);
            }
        }
        printf("%d\n",BFS());
    }
}

POJ - 1062 Expensive bride price (BFS solution)

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