2638: 黑白染色

题目链接

题目大意：黑白矩阵，每次可以选择一个四联通块染色，求最少操作次数

题解：Orz cls，ccz大爷题解

我的收获：bfs树

#include <queue>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int MAXN = 54;const int dx[] = { 1, 0, -1, 0 };const int dy[] = { 0, 1, 0, -1 };char color[MAXN * MAXN];vector<int> e[MAXN * MAXN];char buf[MAXN][MAXN];int tag[MAXN][MAXN];void floodfill(int x, int y, int z, char c) {    if (buf[x][y] != c) {        return;    }    buf[x][y] = '\0';    tag[x][y] = z;    for (int i = 0; i < 4; ++i) {        floodfill(x + dx[i], y + dy[i], z, c);    }}int gao(int n, int s) {    int ret = -1;    queue<int> q;    vector<int> d(n, -1);    d[s] = 0;    q.push(s);    while (!q.empty()) {        s = q.front();        q.pop();        ret=d[s];        if (color[s] == 'W') ret--;        for (vector<int>::iterator it = e[s].begin(); it != e[s].end(); ++it) {            if (d[*it] == -1) {                d[*it] = d[s] + 1;                q.push(*it);            }        }    }    return ret;}int main() {    int n, r, c, ans;    scanf("%d%d", &r, &c);    for (int i = 1; i <= r; ++i) {        scanf("%s", buf[i] + 1);    }    n = 0;    for (int i = 1; i <= r; ++i) {        for (int j = 1; j <= c; ++j) {            if (buf[i][j] != '\0') {                color[n] = buf[i][j];                floodfill(i, j, n, buf[i][j]);                ++n;            }        }    }    for (int i = 1; i <= r; ++i) {        for (int j = 1; j <= c; ++j) {            for (int k = 0; k < 4; ++k) {                int x = i + dx[k];                int y = j + dy[k];                if (1 <= x && x <= r && 1 <= y && y <= c && tag[i][j] != tag[x][y]) {                    e[tag[i][j]].push_back(tag[x][y]);                }            }        }    }    for (int i = 0; i < n; ++i) {//去重         sort(e[i].begin(), e[i].end());        e[i].erase(unique(e[i].begin(), e[i].end()), e[i].end());    }    ans = n;    for (int i = 0; i < n; ++i) {        ans = min(ans, gao(n, i));    }    printf("%d\n", ans + 1);    return 0;}