kuangbin背包九讲

HDU - 2602 Bone Collector

思路

01背包板子题

代码

// 二维数组
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 1010;
int f[N][N];
int w[N], v[N];

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &w[i]);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &v[i]);

        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i ++) {
            for (int j = 0; j <= m; j ++) {
                f[i][j] = f[i - 1][j];
                if(j >= v[i])
                    f[i][j] = max(f[i - 1][j], f[i - 1][j - v[i]] + w[i]);
            }
        }

        printf("%d\n", f[n][m]);
    }

    return 0;
}

// 一维数组
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 1010;
int f[N];
int w[N], v[N];

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &w[i]);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &v[i]);

        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i ++) {
            for (int j = m; j >= v[i]; j --) {
                    f[j] = max(f[j], f[j - v[i]] + w[i]);
            }
        }

        printf("%d\n", f[m]);
    }

    return 0;
}

// 打印路径
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 1010;
int f[N][N];
int w[N], v[N];
int path[N][N];

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        int n, m;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &w[i]);
        for (int i = 1; i <= n; i ++)
            scanf("%d", &v[i]);

        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i ++) {
            for (int j = 0; j <= m; j ++) {
                f[i][j] = f[i - 1][j];
                if(j >= v[i]){
                    if(f[i - 1][j - v[i]] + w[i] > f[i][j]) {
                        path[i][j] = 1;
                    }
                    f[i][j] = max(f[i - 1][j], f[i - 1][j - v[i]] + w[i]);
                }
                    
            }
        }

        printf("%d\n", f[n][m]);
        for (int i = n, j = m; i > 0, j > 0; i --) {
            if(path[i][j]) {
                printf("%d ", i);
                j -= v[i];
            }
        }
    }

    return 0;
}

HDU - 2955 Robberies

思路

状态表示$f[i][j]$：考虑前$i$件物品,抢劫$j$元的生存概率
状态计算：$f[i][j]=max(f[i-1][j],f[i-1][j-w[i]\ast p)$
初始化：
memset(f, 0, sizeof f), f[0] = 1.0
答案：从取所以物品价值sum逆序判断是否生存概率大于P, 是则return

代码

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 10010;

int n, w[N], sum;   
double f[N], P, p[N];
const double eps = 1e-9;

int solve() {
    memset(f, 0, sizeof f);
    f[0] = 1.0;
    for (int i = 1; i <= n; i ++)
        for (int j = sum; j >= w[i]; j --) {
            f[j] = max(f[j], f[j - w[i]] * p[i]);
        }

    for (int i = sum; i >= 0; i --)
        if(f[i] - P >= eps)
            return i;
    return 0;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        cin >> P >> n;
        sum = 0, P = 1.0 - P;
        for (int i = 1; i <= n; i ++) {
            cin >> w[i] >> p[i];
            sum += w[i], p[i] = 1.0 - p[i]; // 不被抓的概率
        }

        cout << solve() << endl;
    }

    return 0;
}

CodeForces - 366C

思路

DP + 思维
题目要求取出的物品$\sum a[i]/\sum b[i]=k$
则可以将每个物品的体积看作 $a[i]-k\ast b[i]$
因为这个体积有正有负
则答案必须体积 正负抵消为0 才满足题目要求
因此跑一遍01背包, 最后枚举体积求出答案即可

代码

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 100010;

int dp1[N], dp2[N];
int a[N], b[N], c[N];

int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= n; i ++) {
		scanf("%d", &b[i]);
	}
	memset(dp1, -0x3f, sizeof dp1);
	memset(dp2, -0x3f, sizeof dp2);
	dp1[0] = 0, dp2[0] = 0;

	for (int i = 1; i <= n; i ++) {
		c[i] = a[i] - k * b[i];
	}

	for (int i = 1; i <= n; i ++)
		if(c[i] >= 0) {
			for (int j = 10000; j >= c[i]; j --) {
				dp1[j] = max(dp1[j], dp1[j - c[i]] + a[i]);
			}
		}
		else {
			c[i] = -c[i];
			for (int j = 10000; j >= c[i]; j --) {
				dp2[j] = max(dp2[j], dp2[j - c[i]] + a[i]);
			}
		}

	int ans = -1;
	for (int i = 0; i <= 10000; i ++) {
		if(dp1[i] == 0 && dp2[i] == 0)
			continue;
		ans = max(ans, dp1[i] + dp2[i]);
	}

	printf("%d\n", ans);

	return 0;
}