POJ 1163 The Triangle
2017-02-06 21:49
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The Triangle
Description
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
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
Sample Output
Source
IOI 1994
一道简单的动态规划题目,状态转移方程为:dp[i][j] = max(dp[i - 1][j], dp[i-1][j-1])+a[i][j];
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int a[105][105],dp[105][105];
int main(){
int n;
while(~scanf("%d",&n)){
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= i; j ++)
scanf("%d",&a[i][j]);
}
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]) + a[i][j];
}
}
int Max = 0;
for(int i = 1; i <= n; i ++)
Max = max(Max,dp
[i]);
printf("%d\n",Max);
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 47093 | Accepted: 28499 |
7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
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
IOI 1994
一道简单的动态规划题目,状态转移方程为:dp[i][j] = max(dp[i - 1][j], dp[i-1][j-1])+a[i][j];
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int a[105][105],dp[105][105];
int main(){
int n;
while(~scanf("%d",&n)){
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= i; j ++)
scanf("%d",&a[i][j]);
}
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]) + a[i][j];
}
}
int Max = 0;
for(int i = 1; i <= n; i ++)
Max = max(Max,dp
[i]);
printf("%d\n",Max);
}
return 0;
}
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