hdu - 4308 - Saving Princess claire - 简单宽搜

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本文介绍了解决HDU 4308问题的方法，该问题是求从起点Y到终点C的最少花费路径，路径中不可通过障碍物#，经过*会增加成本，所有P点必须被联通。采用宽度优先搜索算法，首次遇到P点时将所有相连的P点加入队列。

题意：http://acm.hdu.edu.cn/showproblem.php?pid=4308

从Y到C，遇#不能走，遇*花钱，全部的P连通，问最少花费。

解：

简单宽搜。第一次遇到p就将所有的连通的p都放进队列。

View Code 

#include<cstdio>
#include<queue>
#include<cstring>

using namespace std;
struct point{
    int x, y;
    int sum;
};
bool flag[5000 + 10][5000 + 10];
int r, c, cost;
char map[5000 + 10][5000 + 10];
int dir[4][2] = {
, 0,
    -1, 0,
, 1,
, -1
};
int count_;
int xx[5000 + 10];
int yy[5000 + 10];
int bfs(point start, point end){
    queue < point > q;
    q.push(start);
    int ans = -1;
    int flag1 = 0;
    while(!q.empty()){
        point tmp = q.front();
        q.pop();
        for(int i = 0; i < 4; i ++){
            int x = tmp.x + dir[i][0];
            int y = tmp.y + dir[i][1];
            if(x < 0 || x >= r || y < 0 || y >= c){
                continue;
            }
            if(x == end.x && y == end.y){
                //return tmp.sum;
                if(tmp.sum < ans || ans == -1)
                    ans = tmp.sum;
                continue;
            }
            if(map[x][y] == '#')
              continue;
            if(flag[x][y] == 1)
              continue;
            point now;
            now.x = x;
            now.y = y;
            now.sum = tmp.sum;
            if(map[x][y] == '*'){
                flag[x][y] = 1;
                now.sum = tmp.sum + cost;
                q.push(now);
            }
            if(map[x][y] == 'P'){/*
                if(y + 2 < c && map[x][y + 2] == 'P'){
                    flag[x][y + 2] = 1;
                    now.y = y + 2;
                    q.push(now);
                }
                if(y - 2 >= 0 && map[x][y - 2] == 'P'){
                    flag[x][y - 2] = 1;
                    now.y = y - 2;
                    q.push(now);
                }
                if(x + 2 < r && map[x + 2][y] == 'P'){
                    flag[x + 2][y] = 1;
                    now.x = x + 2;
                    q.push(now);
                }
                if(x - 2 >= 0 && map[x - 2][y] == 'P'){
                    flag[x - 2][y] = 1;
                    now.x = x - 2;
                    q.push(now);
                }*/
                if(flag1 == 1) continue;
                flag1 = 1;
                for(int j = 0; j < count_; j ++){
                    now.x = xx[j];
                    now.y = yy[j];
                    flag[now.x][now.y] = 1;
                    q.push(now);
                }
            }
        }
    }
    //return -1;
    return ans;
}
int main(){
    while(~scanf("%d%d%d", &r, &c, &cost)){
        for(int i = 0; i < r; i ++){
            scanf("%s", map[i]);
        }
        point start, end;
        memset(flag, 0, sizeof(flag));
        count_ = 0;
        for(int i = 0; i < r; i ++){
            for(int j = 0; j < c; j ++){
                if(map[i][j] == 'Y'){
                    start.x = i;
                    start.y = j;
                    flag[i][j] = 1;
                }
                if(map[i][j] == 'C'){
                    end.x = i;
                    end.y = j;
                }
                if(map[i][j] == 'P'){
                    xx[count_] = i;
                    yy[count_ ++] = j;
                }
            }
        }
        start.sum = 0;
        int f = bfs(start, end);
        if(f == -1)
          puts("Damn teoy!");
        else
          printf("%d\n", f);
    }
    return 0;
}