The Triangle(POJ-1163)
2017-09-27 20:18
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The Triangle
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 51473 Accepted: 31121
Description
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
题目的POJ链接:http://poj.org/problem?id=1163
这是我第一个完全自主做出来DP题,虽然这道题挺水的,但我还是挺开心哒~
题意:求一条权值最大的路径,要求从三角形顶部开始,每步可以向左下或右下走,直到三角形的底部
DP的话,可以从上到下也可以从下到上,状态转移方程为:
dp[i][j] = max ( dp[i-1][j] + dp[i][j], dp[i-1][j-1] + dp[i][j] ) ;
AC的代码
更新
若从下向上DP会更简练一些,状态转移方程为:
dp[i][j] += max(dp[i+1][j], dp[i+1][j+1]);
AC的代码
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 51473 Accepted: 31121
Description
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
题目的POJ链接:http://poj.org/problem?id=1163
这是我第一个完全自主做出来DP题,虽然这道题挺水的,但我还是挺开心哒~
题意:求一条权值最大的路径,要求从三角形顶部开始,每步可以向左下或右下走,直到三角形的底部
DP的话,可以从上到下也可以从下到上,状态转移方程为:
dp[i][j] = max ( dp[i-1][j] + dp[i][j], dp[i-1][j-1] + dp[i][j] ) ;
AC的代码
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 110 int dp ; int my_max(int, int); int main(void) { int i,j,k,n; scanf("%d",&n); memset(dp, 0, sizeof(dp)); for(i = 1; i <= n; i++) for(j = 1; j <= i; j++) scanf("%d",&dp[i][j]); for(i = 1; i <= n; i++) for(j = 1; j <=i; j++) dp[i][j] = my_max(dp[i-1][j]+dp[i][j], dp[i-1][j-1]+dp[i][j]); k = 0; for(i = 0; i <= n; i++) if(dp [i] > k) k = dp [i]; printf("%d\n", k); return 0; } int my_max(int a, int b) { return a>b?a:b; }
更新
若从下向上DP会更简练一些,状态转移方程为:
dp[i][j] += max(dp[i+1][j], dp[i+1][j+1]);
AC的代码
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <vector> #include <stack> #include <queue> using namespace std; #define N 110 int dp ; int max(int,int); int main(void) { int i,j,k,n; scanf("%d",&n); for(i = 1; i <= n; i++) for(j = 1; j <= i; j++) scanf("%d",&dp[i][j]); for(i = n-1; i >=1; i--) for(j = i; j >=1; j--) dp[i][j] += max(dp[i+1][j], dp[i+1][j+1]); printf("%d\n", dp[1][1]); return 0; } int max(int a, int b) { return a>b?a:b; }
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