題目連結:https://www.nowcoder.com/acm/contest/118/A

本來想用記憶化搜尋的，搞了半天弄不出來，乾脆跑dijkstra搞搞吧。

#include<bits/stdc++.h>
using namespace std;

#define rep(i,j,k) for(int i=j;i<=k;i++)
typedef long long ll;
typedef pair<int,int> P;

struct edge{
    int v;
    int val;
};
int sx,sy,ex,ey;
int Map[205][205];
vector<edge> G[20005];
int d[20005];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int n;

void dijkstra(int st){
    for(int i=0;i<=(n+1)*n;i++) d[i]=9999999;
    d[st]=0;
    priority_queue<P,vector<P>, greater<P> > q;
    q.push(P(0,st));
    while(!q.empty()){
        P tmp=q.top();
        q.pop();
        if(tmp.first>d[tmp.second]) continue;
        for(int i=0;i<G[tmp.second].size();i++){
            if(d[G[tmp.second][i].v]>tmp.first+G[tmp.second][i].val){
                d[G[tmp.second][i].v]=tmp.first+G[tmp.second][i].val;
                q.push(P(d[G[tmp.second][i].v],G[tmp.second][i].v));
            }
        }
    }
}

int main(){
    while(cin>>n){
        rep(i,1,n*n+n) G[i].clear();
        rep(i,1,n) rep(j,1,n) {
            scanf("%d",&Map[i][j]);
            if(Map[i][j]==-1){
                sx=i,sy=j;
                Map[i][j]=0;
            }
            if(Map[i][j]==-2){
                ex=i;
                ey=j;
                Map[i][j]=0;
            }
        }
        rep(i,1,n) rep(j,1,n){
            for(int k=0;k<4;k++){
                int xx=i+dx[k];
                int yy=j+dy[k];
                if(xx<=0||yy<=0||xx>n||yy>n) continue;
                G[i*n+j].push_back(edge{xx*n+yy,Map[xx][yy]});
            }
        }

        dijkstra(sx*n+sy);

        cout<<d[ex*n+ey]<<endl;

    }
    return 0;
}