记录编号 |
469375 |
评测结果 |
AAAAAAAAAAAAAAAAAAAA |
题目名称 |
[NOIP 2015]斗地主 |
最终得分 |
100 |
用户昵称 |
panda_2134 |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.693 s |
提交时间 |
2017-11-03 07:54:47 |
内存使用 |
0.32 MiB |
显示代码纯文本
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, w[] = {14, 12, 13, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 }; //权重
int T, N, Ans, cd[20];
void Init() {
int t = 0, c = 0;
for (int i = 1; i <= N; i++) {
scanf("%d%d", &t, &c);
cd[w[t]]++;
}
}
/*
单独估计打单牌的代价
总结:对于暴搜类问题,当搜索对象先后顺序无关紧要时,
如果某几类分支的限制条件过少,导致分支众多,就考虑
能否一次性地合并处理,把多个分支变为一个估价函数,
也就是在靠近答案时一并计算贡献。
*/
inline int h() {
int ret=0;
for (int i = 1; i <= 14; i++) {
int t = cd[i];
ret += t/4; t %= 4;
ret += (bool)t;
}
return ret;
}
void DFS(int st) {
if (st >= Ans) return;
Ans=min(Ans, st+h()); //打单牌/全打完
//四带二-1
for (int i = 1; i <= 14; i++) {
if (cd[i] >= 4) {
cd[i] -= 4;
for (int j = 1; j <= 14; j++) if (cd[j]) {
cd[j]--;
for (int k = 1; k <= 14; k++) if (cd[k]) {
cd[k]--;
DFS(st + 1);
cd[k]++;
}
cd[j]++;
}
cd[i] += 4;
}
}
//四带二-2
for (int i = 1; i <= 14; i++) {
if (cd[i] >= 4) {
cd[i] -= 4;
for (int j = 1; j <= 14; j++) if (cd[j] >= 2) {
cd[j]-=2;
for (int k = 1; k <= 14; k++) if (cd[k] >= 2) {
cd[k]-=2;
DFS(st + 1);
cd[k]+=2;
}
cd[j]+=2;
}
cd[i] += 4;
}
}
//三顺子 不包括13,14
for (int i = 1; i <= 12; i++) {
int j;
for (j = i; cd[j] >= 3 && j <= 12; j++); //[i,j)
if (j - i >= 2) {
for (int k = i; k < j; k++) cd[k] -= 3;
DFS(st + 1);
for (int k = i; k < j; k++) cd[k] += 3;
}
}
//双顺子
for (int i = 1; i <= 12; i++) {
int j;
for (j = i; cd[j] >= 2 && j <= 12; j++);
if (j - i >= 3) {
for (int k = i; k < j; k++) cd[k] -= 2;
DFS(st + 1);
for (int k = i; k < j; k++) cd[k] += 2;
}
}
//单顺子
for (int i = 1; i <= 12; i++) {
int j;
for (j = i; cd[j] >= 1 && j <= 12; j++);
if (j - i >= 5) {
for (int k = i; k < j; k++) cd[k]--;
DFS(st + 1);
for (int k = i; k < j; k++) cd[k]++;
}
}
//三带二
for (int i = 1; i <= 14; i++) if (cd[i] >= 3) {
cd[i] -= 3;
for (int j = 1; j <= 14; j++) if (cd[j] >= 2) {
cd[j] -= 2;
DFS(st + 1);
cd[j] += 2;
}
cd[i] += 3;
}
//三带一
for (int i = 1; i <= 14; i++) if (cd[i] >= 3) {
cd[i] -= 3;
for (int j = 1; j <= 14; j++) if (cd[j] >= 1) {
cd[j]--;
DFS(st + 1);
cd[j]++;
}
cd[i] += 3;
}
}
void Work() {
//预处理上界
//贪心打牌
Ans=h();
DFS(0);
printf("%d", Ans);
if (T) putchar(10);
}
void Clear() {
memset(cd, 0, sizeof(cd)); Ans = 0;
}
int main() {
freopen("landlords.in", "r", stdin);
#ifndef DEBUG
freopen("landlords.out", "w", stdout);
#endif
scanf("%d%d", &T, &N);
while (T--) {
Init(); Work(); Clear();
}
return 0;
}