河南第五届ACM省赛（遥 控 器）（BFS+优先队列）

题目地址：http://acm.nyist.net/JudgeOnline/problem.php?pid=543

思路：队友教的我，能想到广搜，队友用了优先队列并且用visit数组标记已经到达的频道号

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>

const int inf = 0x7f7f7f7f;//2139062143
typedef long long ll;
using namespace std;

int a[15];
int visit[100];//标记已经走过的频道号
int start,end1;

struct node
{
    int loc;
    int step;
    friend bool operator < (node a,node b)
    {
        return a.step > b.step;//最小值优先
    }
};

int bfs()
{
    priority_queue<node> pq;
    node now,next;
    now.loc = start;
    now.step = 0;
    visit[start] = 1;
    pq.push(now);
    while(!pq.empty())
    {
        next = pq.top();
        pq.pop();
        if(next.loc == end1)
        {
            return next.step;
        }
        if(a[11])//频道号增加1
        {
            now.loc = (next.loc + 1) % 100;
            now.step = next.step + 1;
            if(!visit[now.loc])
            {
                visit[now.loc] = 1;
                pq.push(now);
            }
        }
        if(a[12])//频道号减少1
        {
            now.loc = (next.loc - 1 + 100) % 100;
            now.step = next.step + 1;
            if(!visit[now.loc])
            {
                visit[now.loc] = 1;
                pq.push(now);
            }
        }
        for(int i=0; i<10; i++)//到0-9的频道号
        {
            if(a[i] && !visit[i])
            {
                now.loc = i;
                now.step = next.step + 1;
                visit[now.loc] = 1;
                pq.push(now);
            }
        }
        for(int i=10; i<100; i++)//到10-99的频道号
        {
            if(a[10] && a[i/10] && a[i%10] && !visit[i])
            {
                now.loc = i;
                now.step = next.step + 3;
                visit[now.loc] = 1;
                pq.push(now);
            }
        }
    }
    return -1;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(visit,0,sizeof(visit));
        scanf("%d%d%d%d",&a[1],&a[2],&a[3],&a[11]);
        scanf("%d%d%d%d",&a[4],&a[5],&a[6],&a[12]);
        scanf("%d%d%d",&a[7],&a[8],&a[9]);
        scanf("%d%d",&a[10],&a[0]);
        scanf("%d%d",&start,&end1);
        int sum = bfs();
        printf("%d\n",sum);
    }
    return 0;
}

标称：（模拟的太好了）

 
#include <stdio.h>
#include <stdlib.h>
int c[128];
int id(int x)
{
	if(x < 0)
		return x+100;
	else
		return x%100;
}
int to(int x)
{
	x = id(x);
	if(x/10==0)
		return c[x];
	int a = x/10,b = x%10;
	if(!c['-'] || !c[a] || !c[b])
		return 0;
	else
		return 3;
}
int min(int a,int b)
{
	if(a==-1)
		return b;
	else
		return a<b ? a : b;
}
int main()
{
	int z,a,b,ans;
	scanf("%d",&z);
	while(z--)
	{
		ans = -1;
		for(int i=1;i<=3;i++)
			scanf("%d",&c[i]);
		scanf("%d",&c['s']);
		for(int i=4;i<=6;i++)
			scanf("%d",&c[i]);
		scanf("%d",&c['x']);
		for(int i=7;i<=9;i++)
			scanf("%d",&c[i]);
		scanf("%d%d",&c['-'],&c[0]);
		scanf("%d%d",&a,&b);
		for(int i=1;i<100;i++)
		{
			if(c['s'])
			{
				if(id(b-i)==a)
					ans = min(ans,i);
				else if(to(b-i))
					ans = min(ans,to(b-i)+i);
			}
			if(c['x'])
			{
				if(id(b+i)==a)
					ans = min(ans,i);
				else if(to(b+i))
					ans = min(ans,to(b+i)+i);		
			}
		}
		printf("%d\n",ans);
	}
//	system("pause");
	return 0;
}