CF-48E. Ivan the Fool VS Gorynych the Dragon(bfs+dfs+判重回路)

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原创最新推荐文章于 2026-06-12 10:56:35 发布 992 阅读 AI 写同款· GEO 优化 ›

全局最小割一个无向连通网络，去掉一个边集可以使其变成两个连通分量则这个边集就是割集；最小割集当然就权和最小的割集。可以用最小切割最大流定理： 1.min=MAXINT,确定一个源点 2.枚举汇点 3.计算最大流，并确定当前源汇的最小割集，若比min小更新min 4.转到2直到枚举完毕 5.min即为所求输出min 不难看出复杂度很高：枚举汇点要O(n)，最阅读详情

Once upon a time in a kingdom far, far away… Okay, let’s start at the point where Ivan the Fool met Gorynych the Dragon. Ivan took out his magic sword and the battle began. First Gorynych had h heads and t tails. With each strike of the sword Ivan can either cut off several heads (from 1 to n, but not more than Gorynych has at the moment), or several tails (from 1 to m, but not more than Gorynych has at the moment). At the same time, horrible though it seems, Gorynych the Dragon can also grow new heads and tails. And the number of growing heads and tails is determined uniquely by the number of heads or tails cut by the current strike. When the total number of heads and tails exceeds R, Gorynych the Dragon strikes its final blow and destroys Ivan the Fool. That’s why Ivan aims to cut off all the dragon’s heads and tails as quickly as possible and win. The events can also develop in a third way: neither of the opponents can win over the other one and they will continue fighting forever.

The tale goes like this; easy to say, hard to do. Your task is to write a program that will determine the battle’s outcome. Consider that Ivan strikes consecutively. After each blow Gorynych grows a number of new heads and tails depending on the number of cut ones. Gorynych the Dragon is defeated if after the blow he loses all his heads and tails and can’t grow new ones. Ivan fights in the optimal way (fools are lucky), i.e.

if Ivan can win, he wins having struck the least number of blows;
if it is impossible to defeat Gorynych, but is possible to resist him for an infinitely long period of time, then that’s the strategy Ivan chooses;
if Gorynych wins in any case, Ivan aims to resist him for as long as possible.

Input

The first line contains three integers h, t and R (0 ≤ h, t, R ≤ 200, 0 < h + t ≤ R) which represent the initial numbers of Gorynych’s heads and tails and the largest total number of heads and tails with which Gorynych the Dragon does not yet attack. The next line contains integer n (1 ≤ n ≤ 200). The next n contain pairs of non-negative numbers "h_i t_i" which represent the number of heads and the number of tails correspondingly, that will grow if Gorynych has i heads (1 ≤ i ≤ n) cut. The next line contains an integer m (1 ≤ m ≤ 200) and then — the description of Gorynych’s behavior when his tails are cut off in the format identical to the one described above. All the numbers in the input file do not exceed 200.

Output

Print "Ivan" (without quotes) in the first line if Ivan wins, or "Zmey" (that means a dragon in Russian) if Gorynych the Dragon wins. In the second line print a single integer which represents the number of blows Ivan makes. If the battle will continue forever, print in the first line "Draw".

Sample test(s)

input
2 2 4
2
1 0
0 1
3
0 1
0 1
0 0

output
Ivan
2

input
2 2 4
1
0 1
1
1 0

output
Draw

input
2 2 5
1
1 1
1
3 0

output
Zmey
2

思路：本题数据最多就两百，因此可以使用广搜出猎人斩龙的最佳路数，如果猎人无法斩龙，那就深搜判断是否组成回路，从而形成永久的对战模式，还是不行那就更新猎人最终成为龙的食物的最大路数。

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int mm=250;
const int oo=1e9;
int vis[mm][mm];
bool yes;
int h,t,r,n,m;
class node
{
  public:int h,t,c;
}hh[mm],tt[mm];
queue<node >q,p;
int dfs(int x,int y)
{ int&ret=vis[x][y];
  int tx,ty;
  if(x+y>r)return 0;
  if(vis[x][y]==-2)
  {
    yes=1;return 0;
  }
  else if(vis[x][y]==-1)
  {
    ret=-2;int dd=-oo,z;
    int bbs;
    bbs=n<x?n:x;
    for(int i=0;i<bbs;i++)
      {
        tx=x+hh[i].h-i-1;ty=y+hh[i].t;
        z=dfs(tx,ty)+1;
        dd=dd>z?dd:z;
      }
      bbs=m<y?m:y;
      for(int i=0;i<bbs;i++)
      {
        tx=x+tt[i].h;ty=y+tt[i].t-i-1;
        z=dfs(tx,ty)+1;
        dd=dd>z?dd:z;
      }
      ret=dd;
  }
  return ret;
}
int main()
{
  while(cin>>h>>t>>r)
  {
    cin>>n;
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n;i++)
    {
      cin>>hh[i].h>>hh[i].t;
    }
    cin>>m;
    for(int j=0;j<m;j++)
    {
      cin>>tt[j].h>>tt[j].t;
    }
    vis[h][t]=1;
     int x,y,zz;
     while(!q.empty())q.pop();
     node z,p;yes=0;
     z.h=h;z.t=t;z.c=0;
     q.push(z);
     while(!q.empty()&&!yes)
     { z=q.front();q.pop();
      int bbs=n<z.h?n:z.h;
      for(int j=0;j<bbs;j++)
      {
        x=z.h+hh[j].h-j-1;y=z.t+hh[j].t;
        if(x+y>r)continue;
        if(!vis[x][y])
        {vis[x][y]=1;
         p.h=x;p.t=y;p.c=z.c+1;
         if(x==0&&y==0)
          { cout<<"Ivan\n"<<p.c<<"\n";yes=1;
            break;
          }
         ///if(x+y>r&&d>z.c){d=z.c;continue;}
         q.push(p);
        }
      }
      bbs=m<z.t?m:z.t;
      for(int j=0;j<bbs;j++)
      {
        x=z.h+tt[j].h;y=z.t+tt[j].t-j-1;
        if(x+y>r)continue;
        if(!vis[x][y])
        {vis[x][y]=1;
         p.h=x;p.t=y;p.c=z.c+1;
         if(x==0&&y==0)
          { cout<<"Ivan\n"<<p.c<<"\n";yes=1;
            break;
          }
         ///if(x+y>r&&d>z.c){d=z.c;continue;}
         q.push(p);
        }
      }
    }
    if(yes)continue;
    memset(vis,-1,sizeof(vis));
    int ans=dfs(h,t);
    if(yes)cout<<"Draw\n";
    else cout<<"Zmey\n"<<ans<<"\n";
  }
}

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