| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14210 | Accepted: 9432 |
Description
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow‘s score is the sum of the numbers of the cows visited along the way. The cow with the highest
score wins that frame. Input
Output
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
数字三角形问题。。可以自底向上坐dp dp[i][j]=ma[i][j]+max(dp[i+1][j],dp[i+1][j+1])
巨水 。。想当初半年前自己懵懵懂懂的刷dp啥都不懂。。哎 真是个悲伤的故事。。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define maxn 360
#define pp pair<int,int>
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n,dp[maxn][maxn],ma[maxn][maxn];
void solve()
{
for(int i=0;i<n;i++)
dp[n-1][i]=ma[n-1][i];
for(int i=n-2;i>=0;i--)
for(int j=0;j<=i;j++)
dp[i][j]=max(ma[i][j]+dp[i+1][j],ma[i][j]+dp[i+1][j+1]);
printf("%d\n",dp[0][0]);
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
for(int j=0;j<=i;j++)
scanf("%d",&ma[i][j]);
memset(dp,0,sizeof(dp));
solve();
}
return 0;
}也可以自顶向下记忆化搜索。。然后状态数组含义都差不多 。。个人觉着搜索比较好写。。。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define maxn 360
#define pp pair<int,int>
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n,dp[maxn][maxn],ma[maxn][maxn];
int dfs(int x,int y)
{
if(x==n-1)return ma[x][y];
if(dp[x][y]>=0)return dp[x][y];
dp[x][y]=0;
dp[x][y]+=(ma[x][y]+max(dfs(x+1,y),dfs(x+1,y+1)));
return dp[x][y];
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
for(int j=0;j<=i;j++)
scanf("%d",&ma[i][j]);
memset(dp,-1,sizeof(dp));
printf("%d\n",dfs(0,0));
}
return 0;
}POJ 3176-Cow Bowling(DP||记忆化搜索)
原文地址：http://blog.csdn.net/qq_16255321/article/details/41174721
