bzoj千题计划225：bzoj2143: 飞飞侠

http://www.lydsy.com/JudgeOnline/problem.php?id=2143

分层图最短路

把能够弹跳的曼哈顿距离看做能量

dp[i][j][k]表示在（i，j）位置，还有能量k的最少花费

弹跳的曼哈顿距离增加1，能量减1

当能量减为0时，花费费用充满能量

#include<queue>
#include<cstdio>
#include<iostream>
 
#define N 151
 
typedef long long LL;
 
const LL inf=1e17;
 
using namespace std;
 
int n,m;
int energy[N][N],cost[N][N];
 
int X[4],Y[4];
 
LL dp[N][N][N<<1];
bool vis[N][N][N<<1];
 
int dx[5]={0,-1,0,1,0};
int dy[5]={0,0,1,0,-1};
 
struct node
{
    int x,y,k;
    LL val;
     
    node(int x_=0,int y_=0,int k_=0,int val_=0) :x(x_),y(y_),k(k_),val(val_) {}
     
    bool operator < (node p) const
    {
        return val>p.val;
    }
     
}now;
 
priority_queue<node>q;
 
void read(int &x)
{
    x=0; int f=1; char c=getchar();
    while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
    while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); }
    x*=f;
}
 
void dijkstra(int e)
{
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            for(int k=0;k<=n+m-2;++k)
            {
                dp[i][j][k]=inf;
                vis[i][j][k]=false;
            }
    while(!q.empty()) q.pop();
    vis[X[e]][Y[e]][0]=true;
    dp[X[e]][Y[e]][energy[X[e]][Y[e]]]=cost[X[e]][Y[e]];
    now.x=X[e];
    now.y=Y[e];
    now.k=energy[now.x][now.y];
    now.val=cost[now.x][now.y];
    q.push(now);
    int sx,sy,nx,ny,k;
    while(!q.empty() && (!vis[X[1]][Y[1]][0] || !vis[X[2]][Y[2]][0] || !vis[X[3]][Y[3]][0]))
    {
        now=q.top();
        q.pop();
        sx=now.x; sy=now.y; k=now.k;
        if(vis[sx][sy][k]) continue;
        vis[sx][sy][k]=true;
        if(now.k)
        {   
            for(int i=0;i<5;++i)
            {
                nx=now.x+dx[i];
                ny=now.y+dy[i];
                if(nx<=0 || nx>n || ny<=0 || ny>m) continue;
                if(dp[sx][sy][k]<dp[nx][ny][k-1])
                {
                    dp[nx][ny][k-1]=dp[sx][sy][k];
                    q.push(node(nx,ny,k-1,dp[nx][ny][k-1]));
                }
            }
        }
        else
        {
            if(dp[sx][sy][0]+cost[sx][sy]<dp[sx][sy][energy[sx][sy]])
            {
                dp[sx][sy][energy[sx][sy]]=dp[sx][sy][0]+cost[sx][sy];
                q.push(node(sx,sy,energy[sx][sy],dp[sx][sy][energy[sx][sy]]));
            }
        }
    }
}
             
 
int main()
{
    int x;
    read(n); read(m);
    for(int i=1;i<=n;++i) 
        for(int j=1;j<=m;++j)
        {
            read(x);
            energy[i][j]=min(x,max(i-1,n-i)+max(j-1,m-j));
        }
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            read(cost[i][j]);
    for(int i=1;i<=3;++i)
        read(X[i]),read(Y[i]);
    LL ans=inf; char pos;
    dijkstra(1);
    LL a1=dp[X[2]][Y[2]][0],a2=dp[X[3]][Y[3]][0];
    dijkstra(2);
    LL b1=dp[X[1]][Y[1]][0],b2=dp[X[3]][Y[3]][0];
    dijkstra(3);
    LL c1=dp[X[1]][Y[1]][0],c2=dp[X[2]][Y[2]][0];
    if(b1+c1<ans) ans=b1+c1,pos='X';
    if(a1+c2<ans) ans=a1+c2,pos='Y';
    if(a2+b2<ans) ans=a2+b2,pos='Z';
    if(ans>=inf) printf("NO");
    else printf("%c
%lld",pos,ans);
    return 0;
}