鬥地主機器出牌核心代碼

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#define _CRT_SECURE_NO_WARNINGS

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<time.h>

#define Arr_Len(arr) (sizeof(arr)/sizeof(arr[0]))

#define maxn 18

/*

1.出牌方式 以-1結束

一對8的出發：8 8 -1

*/

int cod[maxn] = { 0 };

int mycod[maxn];

int chu[maxn];

int rtmp[maxn] = { 0 }, mtmp[maxn] = { 0 };

int mx, mn;

struct LianD

{

int en; //連對結尾的牌號

int du; //最接近的度數

int cnt; //連對的個數

int tem; //臨時存儲的結尾牌號

//複製值函數

void copfun(struct LianD *Ld){

Ld->cnt = cnt;

Ld->du = du;

Ld->en = en;

Ld->tem = tem;

}

//初始化

void init(){

en = du = cnt = tem = 0;

}

};

enum AlgTyNn{

ShunZi = 1,

LianDui

};

int cmp(const void * arg, const void * brg){

int * ar = (int *)arg;

int * br = (int *)brg;

return *ar - *br;

}

void init(){

int i, hopg, cnt = 0;

memset(mycod, 0, sizeof(mycod));

mycod[cnt] = 17;

for (i = 3; i<16; i++)

cod[i] = 4;

cod[16] = 1; //一對王

cod[17] = 1;

srand(time(NULL));

while (cnt<maxn>

hopg = rand() % maxn;

if (hopg>0 && cod[hopg] > 0){

cod[hopg]--;

mycod[++cnt] = hopg;

}

}

qsort(mycod + 1, cnt - 1, sizeof(mycod - 1), cmp);

}

//對方輸出

void inputs(){

int cnt = 0;

memset(chu, 0, sizeof(chu));

while (1){

scanf("%d", &chu[++cnt]);

if (chu[cnt]<0) break;

}

chu[0] = cnt - 1; //最開頭放置紙牌的數目

}

//打印剩餘

void output(){

printf("\n 剩牌系: \n");

//剩餘牌系

int cont = 0;

for (int j = 3; j<maxn>

cont += mtmp[j];

for (int k = 0; k<mtmp>

printf("%d ", j);

}

}

mtmp[0] = cont;

printf("\t剩牌數: [ %d ]", mtmp[0]);

puts("");

}

//計數排序

void connt(int *tmp, const int ss[]){

//memset(tmp ,0 ,sizeof(tmp));

tmp[0] = ss[0];

for (int i = 1; i <= ss[0]; i++)

tmp[ss[i]]++; //計數排序

}

//判斷是否有炸彈或者王炸

void ZhaDan(){

for (int i = 3; i<maxn>

if (mtmp[i] == 4){

printf("%d %d %d %d ", i, i, i, i);

mtmp[i] -= 4;

output();

return;

}

if (mtmp[16] == 1 && mtmp[17] == 1){

printf("%d %d ", 16, 17);

mtmp[16] = mtmp[17] = 0;

output();

return;

}

puts("沒有牌能壓過！");

return;

}

//如果是連對處理方案

void AlgLDOrShZi(int var){

int i = mn + 1, len = mx - mn + 1, kk = 0;

struct LianD Ld, Tmpld;

Ld.init();

//時間複雜度為0(n^2)

for (i = mn + 1; i<15; i++){

Tmpld.init();

Tmpld.tem = i;

for (int j = i; j

if (mtmp[j] >= var){

if (mtmp[j] == var){

Tmpld.du++;

}

Tmpld.cnt++;

}

else{

Tmpld.init();

Tmpld.tem = j + 1;

}

if (Tmpld.cnt == len){

Tmpld.en = Tmpld.tem;

if (Ld.en == 0 || Ld.du<tmpld.du>

Tmpld.copfun(&Ld);

break;

}

}

}

if (Ld.cnt == len){

puts("提示：");

for (i = Ld.en; i<ld.en>

mtmp[i] -= var;

kk = 0;

while (kk++