[CCPC2017杭州] Heathrock

这题意太长了,我懒得复制。

分析

额……其实就是爆搜模拟。但是状态比较多,需要记双方生命值,3个随从的生命,当前场数,魔力,体力,以及一些玄学优化。

  • 自选目标的牌一定会放在回合的较后出,因为其不会累加魔力。
  • 如果剩下的牌全踢脸也踢不死,直接返回。

之后就能跑得飞快。

代码

一开始输出了"YES",淦。

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#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
using namespace std;
const int MAXN=10;
struct Card{
int type;
int damage;
int cost;
int power;

Card(){}
Card(int type,int damage,int cost,int power):type(type),damage(damage),cost(cost),power(power){}
} cards[MAXN];
bool used[MAXN];
int nlen;
int md1,md2,md3;
int mx1,mx2,mx3;

stack<int> ans_cards,ans_target;
bool dfs(int stone,int power,int bl_me,int bl_en,int m1,int m2,int m3,int depth,bool nearend=false){
if(stone<0)return false;
if(depth>=3)return false;
if(bl_en<=0)return true;

if(depth>=1 && stone==10){
if(m1>0)bl_me-=md1;
if(m2>0)bl_me-=md2;
if(m3>0)bl_me-=md3;

if(m1<=0)m1=mx1;
if(m2<=0)m2=mx2;
if(m3<=0)m3=mx3;
}
if(bl_me<=0)return false;

//full attacking to owner cannot kill it.
int fullattack=0;
int fakepower=power;
int unusd_lian=0;
for(int i=0;i<nlen;i++){
if(used[i])continue;
fakepower+=cards[i].power;
if(cards[i].type==1)continue;
unusd_lian++;
fullattack+=cards[i].damage;
}
if(fullattack+fakepower*(unusd_lian)<bl_en)return false;


for(int i=0;i<nlen;i++){
if(used[i])continue;
const Card &card=cards[i];
int reald=card.damage+power;

if(card.cost>stone)continue;

used[i]=1;
if(!nearend && card.type==1){
if(stone-card.cost>=0 && dfs(stone-card.cost,power+card.power,bl_me,bl_en,m1-reald,m2-reald,m3-reald,depth)){
ans_cards.push(i+1);
ans_target.push(-1);
return true;
}
}
if(!nearend && card.type==2){
if(stone-card.cost>=0 && dfs(stone-card.cost,power+card.power,bl_me,bl_en-reald,m1,m2,m3,depth)){
ans_cards.push(i+1);
ans_target.push(-1);
return true;
}
}
if(card.type==3){
if(stone-card.cost>=0 && dfs(stone-card.cost,power+card.power,bl_me,bl_en-reald,m1,m2,m3,depth,1)){
ans_cards.push(i+1);
ans_target.push(0);
return true;
}

if(m1>0 && stone-card.cost>=0 && dfs(stone-card.cost,power+card.power,bl_me,bl_en,m1-reald,m2,m3,depth,1)){
ans_cards.push(i+1);
ans_target.push(1);
return true;
}

if(m2>0 && stone-card.cost>=0 && dfs(stone-card.cost,power+card.power,bl_me,bl_en,m1,m2-reald,m3,depth,1)){
ans_cards.push(i+1);
ans_target.push(2);
return true;
}

if(m3>0 && stone-card.cost>=0 && dfs(stone-card.cost,power+card.power,bl_me,bl_en,m1,m2,m3-reald,depth,1)){
ans_cards.push(i+1);
ans_target.push(3);
return true;
}
}
used[i]=0;
}

if(dfs(10,0,bl_me,bl_en,m1,m2,m3,depth+1)){
ans_cards.push(-1);
ans_target.push(-1);
return true;
}

return false;
}
int mc1,mc2,mc3;
int bl_me,bl_en;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int kase;cin>>kase;
while(kase--){
memset(used,0,sizeof(used));
while(!ans_cards.empty())ans_cards.pop();
while(!ans_target.empty())ans_target.pop();

cin>>nlen;
cin>>bl_me>>bl_en;

cin>>md1>>mc1>>mx1;
cin>>md2>>mc2>>mx2;
cin>>md3>>mc3>>mx3;
for(int i=0;i<nlen;i++){
cards[i].cost=cards[i].damage=cards[i].power=cards[i].type=0;
int type;cin>>type;
if(type==1 || type==2){
int cost,x,y;
cin>>cost>>x>>y;
cards[i].type=type;
cards[i].cost=cost;
cards[i].damage=x;
cards[i].power=y;
}else{
int cost,x;
cin>>cost>>x;
cards[i].type=type;
cards[i].cost=cost;
cards[i].damage=x;
}
}
ans_cards.push(-1);
ans_target.push(-1);
if(dfs(10,0,bl_me,bl_en,mc1,mc2,mc3,0)){
cout<<"Yes\n";
}else{
cout<<"No\n";
continue;
}

int block=0;
while(!ans_cards.empty()){
queue<int> pack_ans,pack_target;
while(!ans_cards.empty() && ans_cards.top()!=-1){
pack_ans.push(ans_cards.top());
ans_cards.pop();
pack_target.push(ans_target.top());
ans_target.pop();
}

if(!ans_cards.empty()){
ans_cards.pop();
ans_target.pop();
}

if(pack_ans.empty())continue;

cout<<pack_ans.size()<<"\n";
while(!pack_ans.empty()){
cout<<pack_ans.front()<<" ";
pack_ans.pop();
}
cout<<"\n";
while(!pack_target.empty()){
cout<<pack_target.front()<<" ";
pack_target.pop();
}
cout<<"\n";
block++;
}
while(block<3){
cout<<"0\n";
block++;
}
}

return 0;
}

[CCPC2017杭州] Heathrock
https://blog.chenc.me/2019/10/23/ccpc2017hangzhou-heathrock/
作者
CC
发布于
2019年10月23日
许可协议