小戴喜欢玩csgo,他有一批非常强的队友,但是小戴不努力,实力很差劲。现在小戴在考完算法100分并且在餐厅吃了一顿令他满意的大餐后,他便准备今晚五排组队,现在他有一批队友可以选择,csgo中有突破手,狙击手,自由人,补枪手的身份,每个人担任不同的职位有不同的胜率,现规定他们车队胜率是每个人担任位置的胜率之平均数,算法课学的很好的小戴一眼就知道该选择什么队友,但他不说,请你为他设计代码。 先输入c【】为小戴担任各个职业胜率,均不能超过50%,依次输入q【n】【】为其队友的各个身份胜率情况,输出小戴选择的队友和他们的最高胜率。使用回溯法,贪心算法,分支限界法编写c语言程序
时间: 2024-03-18 07:44:26 浏览: 11
由于涉及到多种算法,我将为您提供三个不同算法的代码实现供参考。
1. 回溯法实现
回溯法是一种经典的搜索算法,它通过不断地试探和回溯来寻找问题的最优解。在这道题目中,我们可以通过回溯法来枚举所有可能的组合,找出最优解。
代码实现如下:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXN 10
double c[MAXN], q[MAXN][4];
int best_team[5];
double best_rate = 0.0;
void backtrack(int cur, double rate, int team[5]) {
if (cur == 5) {
if (rate > best_rate) {
best_rate = rate;
memcpy(best_team, team, sizeof(best_team));
}
return;
}
for (int i = 0; i < 4; i++) {
team[cur] = i;
backtrack(cur + 1, rate * q[cur][i], team);
}
}
int main() {
int n = 5;
printf("Please input the winning rate of each role for the main player:\n");
for (int i = 0; i < n; i++) {
scanf("%lf", &c[i]);
}
printf("Please input the winning rate of each role for the teammates:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < 4; j++) {
scanf("%lf", &q[i][j]);
}
}
int team[5];
backtrack(0, 1.0, team);
printf("The best teammates are:\n");
for (int i = 0; i < 5; i++) {
printf("%d ", best_team[i] + 1);
}
printf("\nThe winning rate is %.2lf%%\n", best_rate * 100.0);
return 0;
}
```
2. 贪心算法实现
贪心算法是一种局部最优解策略,它通过不断地选择当前最优解来求得全局最优解。在这道题目中,我们可以通过贪心算法来选择每个位置上胜率最高的队友,然后计算车队的最高胜率。
代码实现如下:
```c
#include <stdio.h>
#include <stdlib.h>
#define MAXN 10
double c[MAXN], q[MAXN][4];
int best_team[5];
double best_rate = 0.0;
int main() {
int n = 5;
printf("Please input the winning rate of each role for the main player:\n");
for (int i = 0; i < n; i++) {
scanf("%lf", &c[i]);
}
printf("Please input the winning rate of each role for the teammates:\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j < 4; j++) {
scanf("%lf", &q[i][j]);
}
}
for (int i = 0; i < 5; i++) {
double max_rate = 0.0;
int max_pos = 0;
for (int j = 0; j < 4; j++) {
if (q[i][j] > max_rate) {
max_rate = q[i][j];
max_pos = j;
}
}
best_team[i] = max_pos;
best_rate *= q[i][max_pos];
}
printf("The best teammates are:\n");
for (int i = 0; i < 5; i++) {
printf("%d ", best_team[i] + 1);
}
printf("\nThe winning rate is %.2lf%%\n", best_rate * 100.0);
return 0;
}
```
3. 分支限界法实现
分支限界法是一种寻找最优解的方法,它通过剪枝和扩展节点的策略来寻找最优解。在这道题目中,我们可以通过分支限界法来枚举所有可能的组合,找出最优解。
代码实现如下:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXN 10
double c[MAXN], q[MAXN][4];
int best_team[5];
double best_rate = 0.0;
typedef struct node {
int cur;
double rate;
int team[5];
} Node;
int cmp(const void *a, const void *b) {
double r1 = *((double *)b + 1) / *((double *)b);
double r2 = *((double *)a + 1) / *((double *)a);
return r1 > r2 ? 1 : -1;
}
void branch_and_bound() {
int n = 5;
Node root;
root.cur = 0;
root.rate = 1.0;
for (int i = 0; i < 5; i++) {
root.team[i] = -1;
}
Node queue[100000];
int front = 0, rear = 0;
queue[rear++] = root;
while (front < rear) {
Node node = queue[front++];
if (node.cur == n) {
if (node.rate > best_rate) {
best_rate = node.rate;
memcpy(best_team, node.team, sizeof(best_team));
}
continue;
}
double q_sort[4][2];
for (int i = 0; i < 4; i++) {
q_sort[i][0] = q[node.cur][i];
q_sort[i][1] = c[node.cur];
}
qsort(q_sort, 4, sizeof(q_sort[0]), cmp);
for (int i = 0; i < 4; i++) {
int pos = q_sort[i][1];
if (node.team[pos] != -1) {
continue;
}
Node next;
next.cur = node.cur + 1;
next.rate = node.rate * q_sort[i][0];
memcpy(next.team, node.team, sizeof(next.team));
next.team[pos] = i;
queue[rear++] = next;
}
}
}
int main() {
printf("Please input the winning rate of each role for the main player:\n");
for (int i = 0; i < 5; i++) {
scanf("%lf", &c[i]);
}
printf("Please input the winning rate of each role for the teammates:\n");
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 4; j++) {
scanf("%lf", &q[i][j]);
}
}
branch_and_bound();
printf("The best teammates are:\n");
for (int i = 0; i < 5; i++) {
printf("%d ", best_team[i] + 1);
}
printf("\nThe winning rate is %.2lf%%\n", best_rate * 100.0);
return 0;
}
```