zoukankan      html  css  js  c++  java
  • USACO section 1.5.1 Number Triangles

    1. 动态规划,f[i,j]=Max{f[i+1,j],f[i+1,j+1]}+a[i,j] (1<=i<=n-1,1<=j<=i)复杂度是o(n^2)。

    2. 以下是代码:

    /*
    ID: dollar4
    PROG: numtri
    LANG: C++
    */
    #include <iostream>
    #include <fstream>
    #include <string>
    #include <algorithm>
    #include <cstring>
    
    using namespace std;
    long long num[1010][1010];
    long max(long a, long b)
    {
        if (a > b)
            return a;
        else return b;
    }
    int main()
    {
        ofstream fout ("numtri.out");
        ifstream fin ("numtri.in");
    
        int n, i, j;
        fin >> n;
        for (i = 0; i < n; i++)
            for (j = 0; j <= i; j++)
                fin >> num[i][j];
        for (i = n - 2; i >= 0; i--)
            for (j = 0; j <= i; j++)
                num[i][j] += max(num[i+1][j], num[i+1][j+1]);
        fout << num[0][0] << endl;
        return 0;
    }
    


    3. 第一次用了BFS做,果断超时,下边是BFS代码:

    /*
    ID: dollar4
    PROG: numtri
    LANG: C++
    */
    #include <iostream>
    #include <fstream>
    #include <string>
    #include <algorithm>
    #include <cstring>
    #include <queue>
    
    using namespace std;
    int num[1010][1010], ans[1010][1010];
    int n;
    struct Node
    {
        int x, y;
    } node;
    queue<Node> q;
    void bfs(Node nod)
    {
        q.push(nod);
        Node next1, next2, temp;
        while (!q.empty())
        {
            temp = q.front();
            q.pop();
            if (temp.y + 1 < n)
            {
                next1.x = temp.x + 1;
                next1.y = temp.y;
                next2.x = temp.x + 1;
                next2.y = temp.y + 1;
                q.push(next1);
                q.push(next2);
                if (ans[next1.x][next1.y] < ans[temp.x][temp.y] + num[next1.x][next1.y])
                    ans[next1.x][next1.y] = ans[temp.x][temp.y] + num[next1.x][next1.y];
                if (ans[next2.x][next2.y] = ans[temp.x][temp.y] + num[next2.x][next2.y])
                    ans[next2.x][next2.y] = ans[temp.x][temp.y] + num[next2.x][next2.y];
            }
            else break;
        }
        return;
    }
    
    int main()
    {
        ofstream fout ("numtri.out");
        ifstream fin ("numtri.in");
        memset(num, 0, sizeof(num));
        int i, j;
        fin >> n;
        for (i = 0; i < n; i++)
            for (j = 0; j <= i; j++)
                fin >> num[i][j];
        Node start;
        start.x = 0;
        start.y = 0;
        ans[0][0] = num[0][0];
        bfs(start);
        for (i = 0; i < n; i++)
        {
            for (j = 0; j <= i; j++)
                cout << ans[i][j] << ' ';
            cout << endl;
        }
    
    
        int output = ans[0][0];
        for (i = 0; i < n; i++)
            for (j = 0; j <= i; j++)
            {
                if (output < ans[i][j])
                    output = ans[i][j];
            }
        fout << output << endl;
        return 0;
    }
    

    4. 官方参考代码

    We keep track (in the "best" array) of total for the best path ending in a given column of the triangle. Viewing the input, a path through the triangle always goes down or down and to the right. To process a new row, the best path total ending at a given column is the maximum of the best path total ending at that column or the one to its left, plus the number in the new row at that column. We keep only the best totals for the current row (in "best") and the previous row (in "oldbest").

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include <assert.h>
    
    #define MAXR 1000
    
    int
    max(int a, int b)
    {
    	return a > b ? a : b;
    }
    
    void
    main(void)
    {
    	int best[MAXR], oldbest[MAXR];
    	int i, j, r, n, m;
    	FILE *fin, *fout;
    
    	fin = fopen("numtri.in", "r");
    	assert(fin != NULL);
    	fout = fopen("numtri.out", "w");
    	assert(fout != NULL);
    
    	fscanf(fin, "%d", &r);
    
    	for(i=0; i<MAXR; i++)
    		best[i] = 0;
    
    	for(i=1; i<=r; i++) {
    		memmove(oldbest, best, sizeof oldbest);
    		for(j=0; j<i; j++) {
    			fscanf(fin, "%d", &n);
    			if(j == 0)
    				best[j] = oldbest[j] + n;
    			else
    				best[j] = max(oldbest[j], oldbest[j-1]) + n;
    		}
    	}
    
    	m = 0;
    	for(i=0; i<r; i++)
    		if(best[i] > m)
    			m = best[i];
    
    	fprintf(fout, "%d\n", m);
    	exit(0);
    }


  • 相关阅读:
    如何将已有的本地Git 库推送到远端仓库?
    2017(秋)软工作业: (4)用户体验分析
    微信公众号UX分析—— 学生作业小结
    2017(秋)软工作业: (3)用户体验分析
    珞珈旧时光
    用scp这个命令来通过ssh传输文件
    课堂讨论:分析软件
    2017(秋)软工作业: (2)硬币游戏—— 代码分析与改进
    地铁口的零钱箱:
    Swagger .Net配置
  • 原文地址:https://www.cnblogs.com/dollarzhaole/p/3188911.html
Copyright © 2011-2022 走看看