算法设计作业

1.01背包问题-回溯法

#include<bits/stdc++.h>

using namespace std;

int c, n, cw, cv, bestv;
struct Object {
	int w;
	int v;
	double d;
}Q[100];

bool cmp(Object a, Object b) {
	if (a.d >= b.d) return true;
	else return false;
}


int Bound(int i)
{
	int cleft = c - cw;
	int b = cv;
	while (i < n && Q[i].w <= cleft) {
		cleft -= Q[i].w;
		b += Q[i].v;
		i++;
	}
	if (i < n) b += 1.0 * cleft * Q[i].v / Q[i].w;
	return b;
}


void backtrack(int i)
{
	if (i + 1 > n) { bestv = cv; return; }
	if (cw + Q[i].w <= c) {
		cw += Q[i].w;
		cv += Q[i].v;
		backtrack(i + 1);
		cw -= Q[i].w;
		cv -= Q[i].v;
	}
	if (Bound(i + 1) > bestv)
		backtrack(i + 1);
}

int main()
{
	cout << "请输入背包容量和数量";
	cin >> c >> n;
	for (int i = 0; i < n; i++) {
		cout << "请输入第" << i + 1 << "个物品的重量和价值";
		cin >> Q[i].w >> Q[i].v;
		Q[i].d = 1.0 * Q[i].v / Q[i].w;
	}
	sort(Q, Q + n, cmp);
	cw = 0;
	cv = 0;
	backtrack(0);
	cout << "背包能装下的最大价值为" << bestv << endl;
	return 0;
}

2.装载问题

#include<iostream>
using namespace std;

int n, c, w[100], x[100], r, cw, bestw, bestx[100];

void backtrack(int t)
{
	if (t > n) {
		if (cw > bestw)
		{
			for (int i = 1; i <= n; i++)
				bestx[i] = x[i];
			bestw = cw;
		}
		return;
	}
	r -= w[t];
	if (cw + w[t] <= c)
	{
		x[t] = 1;
		cw += w[t];
		backtrack(t + 1);
		cw -= w[t];
	}
	if (cw + r > bestw)
	{
		x[t] = 0;
		backtrack(t + 1);
	}
	r += w[t];
}

int main()
{
	cout << "请输入集装箱数量：";
	cin >> n;
	cout << "请输入轮船载重量：";
	cin >> c;

	r = 0;
	cout << "请依次输入每个集装箱的重量：";
	for (int i = 1; i <= n; i++) {
		cin >> w[i];
		r += w[i];
	}

	cw = 0;
	bestw = 0;

	backtrack(1);

	cout << "轮船能装载的最大重量为：" << bestw << endl;
	cout << "选择的集装箱编号为：";
	for (int i = 1; i <= n; i++) {
		if (bestx[i] == 1) {
			cout << i << " ";
		}
	}
	cout << endl;
	return 0;
}

3.二分查找

#include<iostream>
using namespace std;

// 二分查找函数，返回目标元素第一次出现的位置
int binarySearch(int a[], int x, int n)
{
    int left = 0;
    int right = n - 1;
    int result = -1;
    while (left <= right)
    {
        int mid = (right + left) / 2;
        if (x == a[mid])
        {
            result = mid;
            // 继续向左查找，看是否有更早出现的目标元素
            right = mid - 1;
        }
        else if (x > a[mid])
            left = mid + 1;
        else
            right = mid - 1;
    }
    return result;
}

int main()
{
    int n, c1[105], c2[105], m;
    cout << "请输入n和要访问的个数";
    cin >> n >> m;
    cout << "请输入" << n << "个数组元素：";
    for (int i = 0; i < n; i++)
    {
        cin >> c1[i];
    }
    cout << "请输入" << m << "个要访问的整数：";
    for (int i = 0; i < m; i++)
    {
        cin >> c2[i];
    }
    for (int i = 0; i < m; i++)
    {
        cout << binarySearch(c1, c2[i], n) << " ";
    }
    return 0;
}

4.01背包-动态规划

#include <iostream>
#include <algorithm>
using namespace std;

int w[50], v[50], p[50][50];

void knapsack(int c, int n) {
    int jMax = min(w[n] - 1, c);
    for (int j = 0; j <= jMax; j++)
        p[n][j] = 0;
    for (int j = w[n]; j <= c; j++)
        p[n][j] = v[n];
    for (int i = n - 1; i >= 1; i--) { 
        jMax = min(w[i] - 1, c);
        for (int j = 0; j <= jMax; j++)
            p[i][j] = p[i + 1][j];
        for (int j = w[i]; j <= c; j++)
            p[i][j] = max(p[i + 1][j], p[i + 1][j - w[i]] + v[i]);
    }
    p[1][c] = p[2][c];
    if (c >= w[1])
        p[1][c] = max(p[1][c], p[2][c - w[1]] + v[1]);
}

int main() {
    int n, c;
    cout << "请输入物品数量 n 和背包容量 c: ";
    cin >> n >> c;
    cout << "请依次输入每个物品的重量和价值: " << endl;
    for (int i = 1; i <= n; i++) {
        cin >> w[i] >> v[i];
    }
    knapsack(c, n);
    cout << "背包能装下的最大价值为: " << p[1][c] << endl;
    return 0;
}