标签：hdu1026

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11710    Accepted Submission(s): 3661
Special Judge

5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.

It takes 13 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
FINISH
It takes 14 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
14s:FIGHT AT (4,5)
FINISH
God please help our poor hero.
FINISH

题解：给定一个地图，让你从左上角走到右下角，只能上下左右四个方向，有的格子有怪兽，移动格子和打怪都要消耗时间，求到达终点所消耗的最短时间和路径。

题解：最短时间可以用优先队列求出来，路径用二维结构体数组来记录。每个节点保存它从哪儿来，该点消耗的时间。

#include <cstdio>
#include <queue>
#include <cstring>
#define maxn 102
using std::priority_queue;

struct Node{
    int x, y, time;
    friend bool operator<(Node a, Node b){
        return a.time > b.time;
    }
};
struct node{
    int x, y, time;
} path[maxn][maxn];
char map[maxn][maxn];
int n, m, mov[][2] = {0, 1, 0, -1, -1, 0, 1, 0};

bool check(int x, int y)
{
    return x >= 0 && y >= 0 && x < n && y < m && map[x][y] != 'X';
}

bool BFS(int& times)
{
    Node now, next;
    now.x = now.y = now.time = 0;
    priority_queue<Node> Q;
    Q.push(now);
    while(!Q.empty()){
        now = Q.top(); Q.pop();
        if(now.x == n - 1 && now.y == m - 1){
            times = now.time; return true;
        }
        for(int i = 0; i < 4; ++i){
            next = now;
            next.x += mov[i][0]; next.y += mov[i][1];
            if(!check(next.x, next.y)) continue;
            ++next.time;
            path[next.x][next.y].x = now.x;
            path[next.x][next.y].y = now.y;
            path[next.x][next.y].time = 0;
            if(map[next.x][next.y] != '.'){
                next.time += map[next.x][next.y] - '0';
                path[next.x][next.y].time = map[next.x][next.y] - '0';
            }        
            map[next.x][next.y] = 'X'; Q.push(next);
        }
    }
    return false;
}

void printPath(int times, int x, int y)
{
    if(times == 0) return;
    times -= path[x][y].time;
    printPath(times - 1, path[x][y].x, path[x][y].y);
    printf("%ds:(%d,%d)->(%d,%d)\n", 
        times++, path[x][y].x, path[x][y].y, x, y);
    while(path[x][y].time--)
        printf("%ds:FIGHT AT (%d,%d)\n", times++, x, y);    
}

int main()
{
    int times, i;
    while(scanf("%d%d", &n, &m) != EOF){
        for(i = 0; i < n; ++i)
            scanf("%s", map[i]);
        memset(path, 0, sizeof(path));
        if(BFS(times)){
            printf("It takes %d seconds to reach the target position, let me show you the way.\n", times);
            printPath(times, n - 1, m - 1);
        }else puts("God please help our poor hero.");
        puts("FINISH");
    }
    return 0;
}

HDU1026 Ignatius and the Princess I 【BFS】+【路径记录】

标签：hdu1026

原文地址：http://blog.csdn.net/chang_mu/article/details/38041745