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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#329656#5071. Check Pattern is GoodWatwareWA 0ms3808kbC++173.5kb2024-02-17 00:18:462024-02-17 00:18:46

Judging History

你现在查看的是最新测评结果

  • [2024-02-17 00:18:46]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3808kb
  • [2024-02-17 00:18:46]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

constexpr int M = 220, E = 40010, S = 1, T = 2, inf = 1000000;

bool flag = true;

char c;
int t, n, m, e, mp[M][M], id[M][M], dep[E], x[M], y[M];
int to[E], nxt[E], val[E], head[E], cur[E], tot;

inline void add(int u, int v, int w) {
    to[tot] = v, val[tot] = w, nxt[tot] = head[u], head[u] = tot++;
}
inline void add(int u, int v) {
    // printf("add %d -> %d\n", u, v);
    add(u, v, 1), add(v, u, 0);
}
inline void adde(int u, int v) {
    // printf("add %d -> %d\n", u, v);
    add(u, v + e, inf), add(v + e, u, 0);
    add(v, u + e, inf), add(u + e, v, 0);
}

inline bool bfs() {
    queue<int> q;
    // for (int i = 1; i <= tot; i++) dep[i] = -1;
    memset(dep, -1, e * sizeof(int));
    dep[S] = 0, q.push(S);
    while (!q.empty()) {
        int p = q.front();
        q.pop();
        for (int i = head[p]; i; i = nxt[i])
            if (val[i] && dep[to[i]] == -1) dep[to[i]] = dep[p] + 1, q.push(to[i]);
    }
    return dep[T] != -1;
}
int dfs(int n, int fl) {
    if (n == T) return fl;
    int lf = fl;
    for (int &i = cur[n]; i; i = nxt[i])
        if (dep[to[i]] == dep[n] + 1 && val[i]) {
            int g = dfs(to[i], min(fl, val[i]));
            lf -= g, val[i] -= g, val[i ^ 1] += g;
            if (!lf) return fl;
        }
    return fl - lf;
}
inline int dinic() {
    int result = 0;
    while (bfs()) memcpy(cur, head, e * sizeof(int)), result += dfs(S, inf);
    return result;
}

inline void solve() {
    assert(scanf("%d%d", &n, &m)), e = (n - 1) * (m - 1);
    tot = id[0][0] = 2;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            assert(scanf(" %c", &c));
            if (c == '?') mp[i][j] = -1;
            else mp[i][j] = ((i + j) & 1) ^ (c == 'W');
        }
    }
    int ans = 0;
    for (int i = 1; i < n; i++)
        for (int j = 1; j < m; j++) id[i][j] = ++id[0][0], x[id[0][0]] = i, y[id[0][0]] = j;
    for (int i = 1; i < n; i++)
        for (int j = 1; j < m; j++) {
            bool e0 = false, e1 = false;
            for (int u = i; u <= i + 1; u++)
                for (int v = j; v <= j + 1; v++)
                    if (mp[u][v] == 1) e1 = true;
                    else if (mp[u][v] == 0) e0 = true;
            add(id[i][j], id[i][j] + e, inf), add(id[i][j] + e, id[i][j], 0);
            if (!e1) add(S, id[i][j]), ans++;
            if (!e0) add(id[i][j] + e, T), ans++;
        }
    e += id[0][0] + 1;
    for (int i = 1; i < n; i++)
        for (int j = 1; j < m; j++) {
            if (i + 1 < n) adde(id[i][j], id[i + 1][j]);
            if (j + 1 < m) adde(id[i][j], id[i][j + 1]);
            if (i + 1 < n && j + 1 < m) adde(id[i][j], id[i + 1][j + 1]);
            if (i + 1 < n && j - 1 >= 1) adde(id[i][j], id[i + 1][j - 1]);
        }
    printf("%d\n", ans - dinic());
    bfs();
    for (int i = head[S]; i; i = nxt[i])
        if (val[i] || dep[to[i]] != -1) {
            for (int u = x[to[i]]; u <= x[to[i]] + 1; u++)
                for (int v = y[to[i]]; v <= y[to[i]] + 1; v++)
                    assert(mp[u][v] == 0 || mp[u][v] == -1), mp[u][v] = 0;
        }
    for (int i = 1; i <= n; i++, printf("\n"))
        for (int j = 1; j <= m; j++) {
            if (mp[i][j] == -1) mp[i][j] = 1;
            mp[i][j] ^= (i + j) & 1;
            putchar(mp[i][j] ? 'W' : 'B');
        }
    memset(head, 0, e * sizeof(int));
}

int main() {
    assert(scanf("%d", &t));
    if (t != 10000) flag = false;
    while (t--) solve();
    return 0;
}

详细

Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3808kb

input:

3
2 2
??
??
3 3
BW?
W?B
?BW
3 3
BW?
W?W
?W?

output:

1
WB
BW
2
BWW
WBB
WBW
4
BWB
WBW
BWB

result:

wrong answer There are 1 check pattern in you output, but you answered 2 (test case 2)