hdu 4308 - Saving Princess claire_

题目：

Saving Princess claire_

题意：

王子救公主 ， 蛋疼的王子希望支付最小的费用

给一张r*c的图，图上有'Y', 'C', '*', '#' 和 'P' ， 各表示：

Y：王子的起始点

C：公主的地点，即终点

*：需要王子支付一定费用才能过 ， 支付费用为cost

#：不能通行

P：传送点，每一个传送点可以无消耗直接通往任意一个其余传送点

普通移动无消耗

求救出公主的最小费用

思路：很简单的广搜 ， 只是多了P，处理法：只要走上了P，就把所有的P相关位置全部压入队列，继续广搜就是了

代码：

//#pragma comment(linker, "/STACK:102400000,102400000")#include "iostream"#include "cstring"#include "algorithm"#include "cmath"#include "cstdio"#include "sstream"#include "queue"#include "vector"#include "string"#include "stack"#include "cstdlib"#include "deque"#include "fstream"#include "map"using namespace std;typedef long long LL;const long long LINF = (long long)1e30;const int INF = 522133279;const int MAXN = 100000+100;#define eps 1e-14const int mod = 100000007;char G[5010][5010];bool vis[5010][5010];int dir[4][2] = {1,0,0,1,-1,0,0,-1};struct pr{    int x;    int y;    LL cost;    pr(int _x = 0 , int _y = 0 , int _cost = 0)    : x(_x) , y(_y) , cost(_cost)    {}    bool operator < (const pr & b)const    {        return cost > b.cost;    }}s,e;vector<pr> place_Of_P;int ok;int r,c,cost;LL minc;LL minll(LL a , LL b){    return a > b ? b : a;}bool border(int x , int y){    return (x >= 0 && x < r) && (y >= 0 && y < c);}void bfs(pr s){    priority_queue<pr>  que;    que.push(s);    pr cur,tmp;    int x,y;    while(!que.empty())    {        cur = que.top();        que.pop();        for(int i = 0 ; i < 4 ; i++)        {            x = cur.x + dir[i][0];            y = cur.y + dir[i][1];            if(border(x,y) && !vis[x][y] && G[x][y] != '#')            {                vis[x][y] = 1;                tmp.x = x;                tmp.y = y;                tmp.cost = (G[x][y] == '*' ? cur.cost + cost: cur.cost);                if(G[x][y] == 'P')                    for(int j = 0 ; j < place_Of_P.size() ; j++)                    {                        que.push(pr(place_Of_P[j].x , place_Of_P[j].y , tmp.cost));                        vis[place_Of_P[j].x][place_Of_P[j].y] = 1;                    }                else if(x == e.x && y == e.y)                {                    minc = minll(minc , tmp.cost);                    ok=1;                    return ;                }                else                    que.push(tmp);            }        }    }}int main(){    //freopen("in","r",stdin);    while(cin >> r >> c >> cost)    {        memset(vis , 0 , sizeof(vis));        ok=0;        place_Of_P.clear();        minc = LINF;        for(int i = 0 ; i < r ; i++)            for(int j = 0 ; j < c ; j++)            {                cin >> G[i][j];                if(G[i][j] == 'Y')                {                    vis[i][j]=1;                    s.x = i;                    s.y = j;                }                else if(G[i][j] == 'C')                {                    e.x = i;                    e.y = j;                }                else if(G[i][j] == 'P')                    place_Of_P.push_back(pr(i,j,0));            }        bfs(s);        if(!ok)            cout << "Damn teoy!" << endl;        else            cout << minc << endl;    }    return 0;}