HDU4198-BFS+优先队列

题目：题目链接

题意：给你一个地图，上面"."代表水，"#"表示陆地，"S"表示船的起始位置，"@"表示桥。我们知道船是不可以走陆地

的，所以只有"."和"@"可以走，但是走桥的时候有一个打开桥的时间d，过桥还需要1分钟，因为每个点可以走多次，所

以要用到优先队列，每次都让步数最小的点出队列结合BFS求最优！求出冲出地图的最小消费时间：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int gcd(int  n,int  m){    if(n<m) swap(n,m);    return n%m==0?m:gcd(m,n%m);}int  lcm(int  n,int  m){    if(n<m) swap(n,m);    return n/gcd(n,m)*m;}#define N 10000007int prime[N];void getPrime();void bash();void wzf();void SG();int QuickMod(int a, int b, int n);#define maxn 505struct node{    int x, y, time;    friend bool operator < (node a, node b)    {        return a.time > b.time;    }};priority_queue<node> q;char mp[maxn][maxn];char vis[maxn][maxn];int h, w, d;int sx, sy;int dir[4][2] = {{0,-1},{-1,0},{0,1},{1,0}};int BFS(){    node now, next;    while(!q.empty())        q.pop();    mem(vis, 0);    now.x = sx;    now.y = sy;    now.time = 0;    q.push(now);    vis[sx][sy] = 1;    while(!q.empty())    {        now = q.top();        q.pop();        for(int i = 0; i < 4; ++i)        {            next.x = now.x + dir[i][0];            next.y = now.y + dir[i][1];            next.time = now.time+1;            if(next.x < 0 || next.x >= h || next.y < 0 || next.y >= w)                return next.time;            if(mp[next.x][next.y] == '#' || vis[next.x][next.y])                continue;            if(mp[next.x][next.y] == '@')            {                next.time += d;                q.push(next);                vis[next.x][next.y] = 1;            }            else if(mp[next.x][next.y] == '.')            {                q.push(next);                vis[next.x][next.y] = 1;            }        }    }    return -1;}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d", &h, &w, &d);        for(int i = 0; i < h; ++i)        {            scanf("%s", mp[i]);            for(int j = 0; j < w; ++j)            {                if(mp[i][j] == 'S')                {                    sx = i;                    sy = j;                }            }        }        int ans = BFS();        printf("%d\n", ans);    }    return 0;}int QuickMod(int  a,int b,int n){    int r = 1;    while(b)    {        if(b&1)            r = (r*a)%n;        a = (a*a)%n;        b >>= 1;    }    return r;}void getPrime(){    memset(prime, 0, sizeof(prime));    prime[0] = 1;    prime[1] = 1;    for(int i = 2; i < N; ++i)    {        if(prime[i] == 0)        {            for(int j = i+i; j < N; j+=i)                prime[j] = 1;        }    }}void bash(int n, int m){    if(n%(m+1) != 0)        printf("1\n");    else        printf("0\n");}void wzf(int n, int m){    if(n > m)        swap(n, m);    int k = m-n;    int a = (k * (1.0 + sqrt(5.0))/2.0);    if(a == n)        printf("0\n");    else        printf("1\n");}int numsg[N];void SG(int n){    int sum = 0;    for(int i=0; i < n; i++)    {        scanf("%d",&numsg[i]);        sum ^= numsg[i];    }    if(sum == 0)        printf("No\n");    else    {        printf("Yes\n");        for(int i = 0; i < n; i++)        {            int s = sum ^ numsg[i];            if(s < numsg[i])                printf("%d %d\n", numsg[i], s);//从有num[i]个石子的堆后剩余s个石子        }    }}