POJ1088滑雪（记忆化搜索与dp）

来源：互联网发布：grub4dos 引导linux 编辑：程序博客网时间：2024/05/17 02:17

记忆化搜索代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int Map[105][105], Step[105][105];int nx[] = {0, 1, 0, -1};int ny[] = {1, 0, -1, 0};int R, C, ans;int dfs(int r, int c){    if(Step[r][c] != -1)        return Step[r][c];    Step[r][c] = 1;    for(int i = 0; i < 4; i++)    {        int tx = r+nx[i];        int ty = c+ny[i];        if(tx >= R || tx < 0 || ty >= C || ty < 0)            continue;        if(Map[tx][ty] >= Map[r][c])            continue;        Step[r][c] = max(Step[r][c], dfs(tx, ty) + 1);    }    return Step[r][c];}int main(){    while(scanf("%d%d", &R, &C) != EOF)    {        memset(Step, -1, sizeof(Step));        for(int i = 0; i < R; i++)            for(int j = 0; j < C; j++)                scanf("%d", &Map[i][j]);        ans = -0x3f3f3f3f;        for(int i = 0; i < R; i++)            for(int j = 0; j < C; j++)                ans = max(ans, dfs(i, j));        cout << ans << endl;    }    return 0;}

dp思路：对高度从小到大进行排序，在i点的时候，遍历0~i-1个点(升序排序，i前面的点的高度一定小于等于i),取相邻点间的大的路径长度

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>using namespace std;struct node{    int r, c, h, dp;}Map[10060];bool cmp(node a, node  b){    return a.h < b.h;}int main(){    int R,C, cnt, ans;    while(scanf("%d%d", &R, &C) != EOF)    {        cnt = 0;        ans = 1;        for(int i = 0; i < R; i++)            for(int j = 0; j < C; j++)            {                scanf("%d", &Map[cnt].h);                Map[cnt].r = i;                Map[cnt].c = j;                Map[cnt++].dp = 1;            }        sort(Map, Map+cnt, cmp);        //cout << cnt << endl;        for(int i = 0; i < cnt; i++)        {            int row1 = Map[i].r;            int column1 = Map[i].c;            for(int j = 0; j < i;j++)            {                int  row2 = Map[j].r;                int column2 = Map[j].c;                if(((row1 == row2 && abs(column1 - column2) == 1) || (column1 == column2 && abs(row1 - row2) == 1)) && Map[i].h > Map[j].h)                {                    Map[i].dp = max(Map[i].dp, Map[j].dp + 1);                    ans = max(ans, Map[i].dp);                }            }        }        cout << ans << endl;    }    return 0;}

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