POJ - 3026 Borg Maze bfs+最小生成树。

http://poj.org/problem?id=3026

题意：给你一个迷宫，里面有 ‘S’起点，‘A’标记，‘#’墙壁，‘ ’空地。求从S出发，经过所有A所需要的最短路。你有一个特殊能力，当走到S或A时可以分身出任意多个人一起走。计算路程时就是所有人的总路程之和。

题解：想一下，是裸的最短路套上bfs。

　　先暴力bfs出各个点之间的距离，存边

　　然后kruskal

#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#include
#include<string.h>
#include<set>
#include<string>
using namespace std;
const int maxn = 3e4;;
int len1, len2, p[maxn], ans;
set<int> s;
struct edge {
    int to, from, w;
    edge(int to=0, int from=0, int w=0) :to(to), from(from), w(w) {}
};
vector e;
bool cmp(edge a, edge b) {
    return a.w < b.w;
}
int f[maxn];
int find(int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
}
void un(int x, int y) {
    int u = find(x), v= find(y);
    f[u] = v;
}
bool same(int x, int y) {
    return find(x) == find(y);
}
string map[55];
int id[55][55];
int vis[55][55];
int dir[4][2] = { 0,1 ,1,0, 0,-1, -1,0 };
struct node {
    int x, y, t;
    node(int x=0, int y=0, int t = 0) :x(x), y(y), t(t) {}
};
void bfs(int x, int y) {
    memset(vis, 0, sizeof(vis));
    queue Q;
    Q.push(node(x, y, 0)); vis[x][y] = 1;
    while (!Q.empty()) {
        node now = Q.front();
        Q.pop();
        for (int i = 0; i < 4; i++) {
            int dx = now.x + dir[i][0];
            int dy = now.y + dir[i][1];
            if (map[dx][dy] == '#'||vis[dx][dy]) continue;
            if (map[dx][dy] == 'A')e.push_back(edge(id[x][y], id[dx][dy], now.t+1));
            Q.push(node(dx, dy, now.t + 1));
            vis[dx][dy] = 1;

        }
    }
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        scanf("%d %d\n", &n,&m);
        for (int i = 0; i < m; i++) {
            getline(cin, map[i]);
        }
        memset(id, 0, sizeof(id));
        e.clear();
        int idx = 1;
        for(int i=0;i)
            for (int j = 0; j < n; j++) {
                if (map[i][j] == 'S')map[i][j] = 'A';
                if (map[i][j] == 'A' ) {
                    id[i][j] = idx++;
                }
            }
        for (int i = 0; i)
            for (int j = 0; j < n; j++) {
                if(id[i][j])bfs(i, j);
            }
        for (int i = 1; i < idx; i++) { f[i] = i; }
    sort(e.begin(), e.end(),cmp);
    int res = 0;
    for (int i = 0; i < e.size(); i++) {
        if (same(e[i].to, e[i].from)) continue;
        un(e[i].to, e[i].from);
        res += e[i].w;
    }
    cout << res << endl;
    
    }
    system("pause");
}

 

转载于:https://www.cnblogs.com/SuuT/p/8776686.html