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分层图最短路（其实就是最短路转移时多记录一维的数据
#include <bits/stdc++.h> using namespace std; #define X first #define Y second typedef double db; typedef long long ll; typedef pair<int,int> P; const int MAXN=105; int n,T,dat[MAXN][MAXN]; ll d[MAXN][MAXN][3]; struct node{int x,y,d,w;}; int dx[]={0,0,1,-1}; int dy[]={1,-1,0,0}; priority_queue<node> q; bool operator < (node a,node b){return a.w>b.w;} int main() { scanf("%d%d",&n,&T); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) scanf("%d",&dat[i][j]); memset(d,0x3f,sizeof(d)); d[1][1][0]=0; q.push(node{1,1,0,0}); while(!q.empty()) { node t=q.top();q.pop(); if(d[t.x][t.y][t.d]<t.w) continue; for(int i=0;i<4;i++) { int fx=t.x+dx[i],fy=t.y+dy[i],cur=(t.d+1)%3; if(fx<1||fy<1||fx>n||fy>n) continue; ll cost=t.w+T+(cur==0?dat[fx][fy]:0); if(d[fx][fy][cur]>cost) d[fx][fy][cur]=cost,q.push(node{fx,fy,cur,cost}); } } printf("%lld",min(d[n][n][0],min(d[n][n][1],d[n][n][2]))); return 0; }