洛谷 P1514 引水入城

题意简述

有一个n*m的矩阵，每个格子都有一个高度，第一行可以放蓄水站，可以向四周比这格高度小的格子送水。问可不可以让第n行都有水，若可以输出最少建几个蓄水站，否则输出第n行最少有几个格子没有水。

题解思路

搜索+动态规划

代码

#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
const int fx[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int n, m, ans;
int d[501], map[501][501];
bool b[501][501], dp[501][501];
struct zb{
    int x, y;
};
queue <zb> qq;
void bfs(int x)
{
    memset(b, 0, sizeof(b));
    b[1][x] = 1;
    zb xx;
    xx.x = 1;
    xx.y = x;
    qq.push(xx);
    while (!qq.empty())
    {
        zb xx, yy;
        xx = qq.front();
        qq.pop();
        for (int i = 0; i < 4; ++i)
        {
            yy = xx;
            yy.x += fx[i][0];
            yy.y += fx[i][1];
            if (yy.x >= 1 && yy.x <= n && yy.y >= 1 && yy.y <= m && !b[yy.x][yy.y] && map[xx.x][xx.y] > map[yy.x][yy.y])
            {
                qq.push(yy);
                b[yy.x][yy.y] = 1;
            }
        }
    }
    for (int i = 1; i <= m; ++i)
        for (int j = i; j <= m; ++j)
            if (b[n][i] && b[n][j])
                dp[i][j] = 1;
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            scanf("%d", &map[i][j]);
    for (int i = 1; i <= m; ++i)
    {
        b[1][i] = 1;
        zb xx;
        xx.x = 1;
        xx.y = i;
        qq.push(xx);
    }
    while (!qq.empty())
    {
        zb xx, yy;
        xx = qq.front();
        qq.pop();
        for (int i = 0; i < 4; ++i)
        {
            yy = xx;
            yy.x += fx[i][0];
            yy.y += fx[i][1];
            if (yy.x >= 1 && yy.x <= n && yy.y >= 1 && yy.y <= m && !b[yy.x][yy.y] && map[xx.x][xx.y] > map[yy.x][yy.y])
            {
                qq.push(yy);
                b[yy.x][yy.y] = 1;
            }
        }
    }
    for (int i = 1; i <= m; ++i)
        if (!b[n][i])
            ans++;
    if (ans != 0)
    {
        printf("0\n%d\n", ans);
        return 0;
    }
    for (int i = 1; i <= m; ++i)
        bfs(i);
    for (int i = 1; i <= m; ++i)
        if (dp[1][i]) d[i] = 1;
        else d[i] = 0x7fffffff;
    for (int i = 1; i <= m; ++i)
        for (int j = 1; j <= i; ++j)
            if (dp[j + 1][i] && d[i] > d[j] + 1)
                d[i] = d[j] + 1;
    printf("1\n%d\n", d[m]);
}

洛谷 P1514 引水入城

原文：https://www.cnblogs.com/xuyixuan/p/9429554.html