题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1004
题意:两个数塔相接,求最大的路径和。
解法:简单DP。
代码:
#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>
using namespace std;
int t, n;
int dp[220][220];
int p[220][220];
int main()
{
scanf("%d",&t);
for (int ca = 1; ca <= t;ca++)
{
scanf("%d",&n);
int tmp = n - 1;
int tmp2 = n - 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
scanf("%d",&p[i][j]);
for (int i = n + 1; i <= 2 * n - 1; i++)
for (int j = 1; j <= 2 * n - i; j++)
scanf("%d", &p[i][j]);
memset(dp,0,sizeof(dp));
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1]) + p[i][j];
for (int i = n + 1; i <= 2 * n - 1; i++)
for (int j = 1; j <= 2 * n - i; j++)
dp[i][j] = max(dp[i-1][j],dp[i-1][j+1])+p[i][j];
printf("Case %d: %d\n",ca,dp[2*n - 1][1]);
}
return 0;
}
时间: 2024-10-15 08:02:21