POJ 1163&& 3176 The Triangle(DP)
The Triangle
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 41169 | Accepted: 24882 |
Description
7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1)
Input
Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle,
all integers, are between 0 and 99.
Output
Your program is to write to standard output. The highest sum is written as an integer.
Sample Input
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
Sample Output
30
Source
题意:求走过这个三角形时的最大数值。起点为第一行的唯一的那一个数,终点是第n行的某一个数。当中要走dp[i][j]的话。他的上一步仅仅能是dp[i-1][j-1]或者dp[i-1][j];
#include<iostream> #include<algorithm> #include<stdio.h> #include<string.h> #include<stdlib.h> #define inf 9999 #define INF -9999 using namespace std; int n; int dp[361][361]; int main() { while(scanf("%d",&n)!=EOF) { memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=1;j<=i;j++) { scanf("%d",&dp[i][j]); } } for(int i=1;i<=n;i++) { for(int j=1;j<=i;j++) { dp[i][j] += max(dp[i-1][j-1],dp[i-1][j]); } } int maxx = 0; for(int i=1;i<=n;i++) { if(maxx<dp[n][i]) { maxx = dp[n][i]; } } printf("%d\n",maxx); } return 0; }