动态规划

来源：互联网发布：mysql数据库的优点编辑：程序博客网时间：2024/06/10 20:12

题意：求从左上角走到右下角所经过的路径上的数字之积末尾含有最少的0

dp求出从左上角到右下角所经过的路径所含2,5最少的路径,同时检测所给数组中是否含有0,输出答案。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string.h>#include <string>using namespace std;const int MAXN = 1000 + 38;const int inf = 1000000000;int mt[MAXN][MAXN][2];int dp[MAXN][MAXN][2];int path[MAXN][MAXN][2];int temp[MAXN][MAXN];int cal2(int x){    int res = 0;    while (x % 2 == 0)    {        res++;        x /= 2;    }    return res;}int cal5(int x){    int res = 0;    while (x % 5 == 0)    {        res++;        x /= 5;    }    return res;}void dfs(int x, int y, int k){    if (x == 1 && y == 1) return ;    if (x < 1 || y < 1) return ;    if (path[x][y][k])    {        dfs(x - 1, y, k);        printf("D");    }    else    {        dfs(x, y - 1, k);        printf("R");    }}void input(){    int n;    while (scanf("%d", &n) != EOF)    {        int flag = 0;        for (int i = 1; i <= n; i++)        {            for (int j = 1; j <= n; j++)            {                scanf("%d", &temp[i][j]);                if (temp[i][j] == 0)                {                    flag = i;                    continue;                }                mt[i][j][0] = cal2(temp[i][j]);                mt[i][j][1] = cal5(temp[i][j]);            }        }        for (int i = 0; i <= n; i++)        {            for (int j = 0; j <= n; j++)            {                dp[i][j][0] = dp[i][j][1] = inf;    //无穷大,因为dp求小值            }        }        dp[1][1][0] = mt[1][1][0], dp[1][1][1] = mt[1][1][1];    //初始化        for (int i = 1; i <= n; i++)        {            for (int j = 1; j <= n; j++)            {                if (temp[i][j] == 0) continue;                if (i == 1 && j == 1) continue;                if (dp[i - 1][j][0] < dp[i][j - 1][0] && i > 1)                {                    dp[i][j][0] = dp[i - 1][j][0] + mt[i][j][0];                    path[i][j][0] = 1;                }                else                {                    dp[i][j][0] = dp[i][j - 1][0] + mt[i][j][0];                    path[i][j][0] = 0;                }                if (dp[i - 1][j][1] < dp[i][j - 1][1] && i > 1)                {                    dp[i][j][1] = dp[i - 1][j][1] + mt[i][j][1];                    path[i][j][1] = 1;                }                else                {                    dp[i][j][1] = dp[i][j - 1][1] + mt[i][j][1];                    path[i][j][1] = 0;                }                //cout << dp[i][j][1] << ' ';            }            //cout << endl;        }        int ans = min(dp[n][n][0], dp[n][n][1]);        int k = dp[n][n][0] < dp[n][n][1] ? 0 : 1;        if (ans > 1 && flag)        {            printf("1\n");            for (int i = 1; i < flag; i++)            {                printf("D");            }            for (int i = 1; i < n; i++)            {                printf("R");            }            for (int i = flag; i < n; i++)            {                printf("D");            }            printf("\n");        }        else        {            printf("%d\n", ans);            dfs(n, n, k);            printf("\n");        }    }}int main(){    input();    return 0;}

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