POJ 3026 Borg Maze 最小生成树

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这道题大意很好懂，有一个S点和不超过100个A点，求这些点构成的完全图的最小生成树。

题意虽简单，但实现略繁琐，而且Input有陷阱。

在n，m这一行行末会有多余空格。。。（无聊。）

然后用了prim来做（代码有点乱）。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <queue>using namespace std;const int maxn = 110;const int maxDot = 110;const int INF = 0x3f3f3f3f;int g[maxn][maxn];int n, m, T;int dis[maxDot][maxDot];bool vis[maxn][maxn];int cost[maxDot];int cnt;struct nd {        int x, y, dpt;} dot[maxDot];struct node {        int x, y, dpt;        node(int a, int b, int c) : x(a), y(b), dpt(c) {}};int dx[4] = { -1, 0, 1, 0 };int dy[4] = { 0, -1, 0, 1 };bool check(int x, int y) {        return x >= 0 && x < n && y >= 0 && y < m && g[x][y] >= 0;}void bfs(int st) {        memset(vis, 0, sizeof(vis));        queue<node> q;        dot[st].dpt = 0;        q.push(node(dot[st].x, dot[st].y, dot[st].dpt));        while (!q.empty()) {                node u = q.front(); q.pop();                int x = u.x, y = u.y;                if (vis[x][y]) continue;                vis[x][y] = 1;                if (g[x][y] >= 1) {                        int ed = g[x][y];                        dis[st][ed] = dis[ed][st] = u.dpt;                }                for (int i = 0; i < 4; i++)                        if (check(x + dx[i], y + dy[i]))                                q.push(node(x + dx[i], y + dy[i], u.dpt + 1));        }}bool used[maxDot];int prim() {        for (int i = 1; i <= cnt; i++) {                cost[i] = INF;                used[i] = 0;        }        cost[1] = 0;        int res = 0;        while (true) {                int v = -1;                for (int u = 1; u <= cnt; u++)                        if (!used[u] && (v == -1 || cost[v] > cost[u]))                                v = u;                if (v == -1) break;                used[v] = 1;                res += cost[v];                for (int u = 1; u <= cnt; u++)                        cost[u] = min(cost[u], dis[v][u]);        }        return res;}int main() {        char s[maxn];        scanf("%d", &T);        while (T--) {                scanf("%d%d", &m, &n);                while ('\n' != getchar()) ;                memset(g, -1, sizeof(g));                cnt = 1;                for (int i = 0; i < n; i++) {                        fgets(s, maxn, stdin);                        int len = strlen(s) - 1;                        for (int j = 0; j < len; j++)                                if (s[j] == '#')                                        g[i][j] = -1;                                else if (s[j] == ' ')                                        g[i][j] = 0;                                else if (s[j] == 'A')                                        g[i][j] = ++cnt;                                else                                        g[i][j] = 1;                }                for (int i = 0; i < n; i++)                        for (int j = 0; j < m; j++)                                if (g[i][j] > 0) {                                        int t = g[i][j];                                        dot[t].x = i;                                        dot[t].y = j;                                }                for (int i = 1; i <= cnt; i++)                        for (int j = 1; j <= cnt; j++)                                dis[i][j] = INF;                for (int i = 1; i <= cnt; i++)                        bfs(i);                for (int i = 1; i <= cnt; i++) dis[i][i] = 0;                printf("%d\n", prim());        }        return 0;}

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