洛谷 1514 引水入城

作为一道提高T2难度的题目，它成功的让我自闭了。。。

读完题第一感觉：flood_fill，第二感觉：第二问不可做。。。

于是思考许久，还是翻开了题解，然后看到一句：“每个蓄水厂覆盖的区间必然连续，否则输出0”。

题目有变的可做起来，于是瞎搞了一波flood_fill + 区间DP。

结果只有80分。。。一个点TLE，一个点WA。。。

TLE好办，对第一行来波记忆化就完了，WA就难办了，下了个数据点发现，还要特判ｎ＝＝１的情况。

终于AC辣！！！！

AC 代码:

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

inline int min(const int a, const int b) { return a > b ? b : a; }
inline int max(const int a, const int b) { return a < b ? b : a; }

inline void read(int & x)
{
    x = 0;
    int k = 1;
    char c = getchar();
    while (!isdigit(c))
        if (c == '-') c = getchar(), k = -1;
        else c = getchar();
    while (isdigit(c))
        x = (x << 1) + (x << 3) + (c ^ 48),
        c = getchar();
    x *= k;
}

struct node
{
    int x, y;
}fir, cur;

int n, m, nx, ny, out;
int h[505][505];
int v[505][505];
int dk[505];
int dp[505][505];
int mem[505][505];
int l[505], r[505], ans[505];
int brd[505];
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

void bfs(int x, int now)
{
    if (dk[now]) 
    {
        l[now] = l[dk[now]],
        r[now] = r[dk[now]],
        ans[now] = ans[dk[now]];
        return;
    }
    std::queue <node> Q;
    fir.x = x, fir.y = now;
    Q.push(fir);
    v[x][now] = 1;
    while (!Q.empty())
    {
        cur = Q.front(); Q.pop();
        for (int i = 0; i < 4; ++i)
        {
            nx = cur.x + dx[i],    ny = cur.y + dy[i];
            if (v[nx][ny] || nx < 1 || nx > n || ny > m || ny < 1 || h[nx][ny] >= h[cur.x][cur.y]) continue;
            fir.x = nx, fir.y = ny;
            if (nx == 1) dk[ny] = now;
            v[nx][ny] = 1; Q.push(fir);
            if (nx == n) l[now] = min(l[now], ny), r[now] = max(r[now], ny), ++ans[now], mem[n][ny] = 1;
        }
    }
}

void flood_fill(int now)
{
    memset(v, false, sizeof(v));
    bfs(1, now); 
    for (int i = l[now]; i <= r[now]; ++i) brd[i] = 1;
    if (l[now] <= r[now])
        for (int i = l[now]; i <= r[now]; ++i)
            for (int j = i; j <= r[now]; ++j)
                dp[i][j] = 1;
}

inline void dp_it()
{
    for (int len = 1; len <= m; ++len) 
        for (int i = 1, j = i + len - 1; j <= m; ++i, j = i + len - 1)
            for (int k = i; k < j; ++k)
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
}

int main()
{
    memset(l, 0x3f, sizeof(l));
    memset(dp, 0x3f, sizeof(dp));
    read(n), read(m); bool flag = true;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            read(h[i][j]);
    for (int i = 1; i <= m; ++i)
        flood_fill(i);
    for (int i = 1; i <= m; ++i) if (!mem[n][i]) ++out; 
    if (out != 0) { if (n == 1) printf("1
%d", out); else printf("0
%d", out); return 0; }
    dp_it();
    printf("1
%d", dp[1][m]);
    return 0;
}