poj - 1088 - 滑雪（dp）

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题意：一个R * C的矩阵(1 <= R,C <= 100)，元素代表该点的高度h（0<=h<=10000），从任意点出发，每次只能走上、下、左、右，且将要到的高度要比原高度小，求最长路。

题目链接：http://poj.org/problem?id=1088

——>>设dp[i][j]表示从ij位置出发的最长路，则状态转移方程为：

dp[x][y] = max(dp[x][y], Dp(nNewX, nNewY) + 1);

时间复杂度：O(R * C)

#include <cstdio>#include <cstring>#include <algorithm>using std::max;const int MAXN = 100 + 10;int R, C;int nHeight[MAXN][MAXN];int dp[MAXN][MAXN];int dx[] = {-1, 1,  0, 0};int dy[] = { 0, 0, -1, 1};void Read(){    for (int i = 1; i <= R; ++i)    {        for (int j = 1; j <= C; ++j)        {            scanf("%d", &nHeight[i][j]);        }    }}int Dp(int x, int y){    if (dp[x][y] != -1)    {        return dp[x][y];    }    int& nRet = dp[x][y];    nRet = 1;    for (int i = 0; i < 4; ++i)    {        int nNewX = x + dx[i];        int nNewY = y + dy[i];        if (nNewX >= 1            && nNewX <= R            && nNewY >= 1            && nNewY <= C            && nHeight[nNewX][nNewY] < nHeight[x][y])        {            nRet = max(nRet, Dp(nNewX, nNewY) + 1);        }    }    return nRet;}void Solve(){    int nRet = 0;    memset(dp, -1, sizeof(dp));    for (int i = 1; i <= R; ++i)    {        for (int j = 1; j <= C; ++j)        {            if (dp[i][j] == -1)            {                Dp(i, j);            }        }    }    for (int i = 1; i <= R; ++i)    {        for (int j = 1; j <= C; ++j)        {            if (dp[i][j] > nRet)            {                nRet = dp[i][j];            }        }    }    printf("%d\n", nRet);}int main(){    while (scanf("%d%d", &R, &C) == 2)    {        Read();        Solve();    }    return 0;}

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