ZROI2018普转提day1t4

传送门

分析

就是飞飞侠这道题......

我们可以将这张图建成好几层，每一层可以向下一层的上下左右无代价移动，而对于每个点如果付b[i][j]的代价就可以走到比它高a[i][j]的层上。我们用这种方式优化了建边，然后跑个最短路就行了。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define ri register int
const long long inf = 1e12+;
const int dx[] = {,-,,,};
const int dy[] = {,,,-,};
inline int ra(){
      int x=;char s=getchar();while(!isdigit(s))s=getchar();
      while(isdigit(s)){x=(x<<)+(x<<)+(s-'');s=getchar();}return x;
}
inline long long ra2(){
      long long x=;char s=getchar();while(!isdigit(s))s=getchar();
      while(isdigit(s)){x=(x<<)+(x<<)+(s-'');s=getchar();}return x;
}
struct node {long long wh;int x,y,z;};
bool operator < (node p,node q){return p.wh>q.wh;}
int n,m,a[][],vis[][][],x[],y[];
priority_queue<node>q;
long long b[][],d[][][],w[];
inline void dij(int s,int t1,int t2){
      node A,B;
      for(ri i=;i<=n;++i)
        for(ri j=;j<=m;++j)
          for(ri k=;k<=n+m;++k)d[i][j][k]=inf,vis[i][j][k]=;
      while(!q.empty())q.pop();
      d[x[s]][y[s]][]=;
      A.wh=,A.x=x[s],A.y=y[s],A.z=;
      q.push(A);
      while(!q.empty()){
          if(vis[x[]][y[]][]&&vis[x[]][y[]][]&&vis[x[]][y[]][])
          break;
          A=q.top();q.pop();
          if(vis[A.x][A.y][A.z])continue;
          vis[A.x][A.y][A.z]=;
          if(A.z>){
            for(ri i=;i<;++i){
                if(dx[i]+A.x<||dx[i]+A.x>n)continue;
                if(dy[i]+A.y<||dy[i]+A.y>m)continue;
                if(d[A.x+dx[i]][A.y+dy[i]][A.z-]>A.wh){
                  d[A.x+dx[i]][A.y+dy[i]][A.z-]=A.wh;
                  B.x=A.x+dx[i],B.y=A.y+dy[i],B.z=A.z-,B.wh=A.wh;
                  q.push(B);
                }
            }
          }else {
            if(d[A.x][A.y][a[A.x][A.y]]>A.wh+b[A.x][A.y]){
                d[A.x][A.y][a[A.x][A.y]]=A.wh+b[A.x][A.y];
                B.x=A.x,B.y=A.y,B.z=a[A.x][A.y],B.wh=A.wh+b[A.x][A.y];
                q.push(B);
            }
          }
      }
      w[t1]+=d[x[t1]][y[t1]][],w[t2]+=d[x[t2]][y[t2]][];
      return;
}
int main(){
      int out;
      n=ra(),m=ra();
      for(ri i=;i<=n;++i)
        for(ri j=;j<=m;++j)
          a[i][j]=ra(),a[i][j]=min(a[i][j],n+m);
      for(ri i=;i<=n;++i)
        for(ri j=;j<=m;++j)
          b[i][j]=ra2();
      for(ri i=;i<;++i)x[i]=ra(),y[i]=ra();
      long long minn=inf;
      dij(,,),dij(,,),dij(,,);
      if(w[]<minn)minn=w[],out=;
      if(w[]<minn)minn=w[],out=;
      if(w[]<minn)minn=w[],out=;
      if(minn>=inf)puts("NO");
        else cout<<char(out+'X')<<endl<<minn<<endl;
      return ;
}