这是往年校赛的一道题，最开始做这道题的时候还没有系统的学习过搜索，用了C语言学的回溯法尝试，毫无疑问的TLE；
学习了DFS之后，自己的剪枝功力不够，又是TLE，最后学了BFS之后，哇，终于做出来了，别提多开心了，然后意识到这道题其实很简单的，剋以用BFS标记法和更改步数法（更改最小消耗），后来发现这种题也可以建图跑迪杰斯特拉做；
BFS标记法：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>using namespace std;
const int N = 100 + 5;
int mat[N][N];
bool visit[N][N];
typedef struct node{int x,y,val;bool operator < (const node x)const {return val > x.val;}
}Node;
const int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};int BFS(int n){priority_queue<Node> Q;Node t;memset(visit,0,sizeof(visit));int x=0,y=0,goalx=n-1,goaly=n-1,newx,newy;t=(Node){x,y,mat[x][y]};visit[x][y] = true;Q.push(t);while(!Q.empty()){t = Q.top();Q.pop();if(t.x == goalx && t.y == goaly) return t.val;for(int d=0;d<4;d++){newx = t.x + dir[d][0];newy = t.y + dir[d][1];if(newx>=0 && newx<n && newy>=0 && newy<n && mat[newx][newy] && !visit[newx][newy]){visit[newx][newy] = true;Q.push((Node){newx,newy,t.val+mat[newx][newy]});}}}return -1;
}void Input_data(int n){for(int i=0;i<n;i++)for(int j=0;j<n;j++){scanf("%d",&mat[i][j]);}
}int main(){int n;while(scanf("%d",&n)==1){Input_data(n);printf("min=%d\n",BFS(n));}
}

BFS更改步数法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>using namespace std;const int N = 100+5;
const int MaxSize = 1e5;
const int INF = (1<<30);
int mat[N][N];
int Min[N][N];
typedef struct node{int x,y,val;bool operator < (const node x) const {return val > x.val;}
}Node;
const int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int BFS(int n){priority_queue<Node> Q;int newx,newy,x=0,y=0,goalx= n-1,goaly=n-1;Node t,s;t.x = x,t.y = y,t.val = mat[0][0];Min[x][y] = mat[x][y];Q.push(t);while(true){t = Q.top();if(t.x == goalx && t.y == goaly ) return t.val;for(int d=0;d<4;d++){newx = t.x+dir[d][0];newy = t.y+dir[d][1];if(newx>=0 && newx<n && newy>=0 && newy<n && mat[newx][newy]){if(t.val + mat[newx][newy] < Min[newx][newy]){s.x = newx,s.y = newy,s.val = t.val+mat[newx][newy];Min[newx][newy] = s.val;Q.push(s);}}}Q.pop();}
}void Init(){for(int i=0;i<N;i++)for(int j=0;j<N;j++){Min[i][j] = INF;}
}void Input_data(int n){for(int i=0;i<n;i++)for(int j=0;j<n;j++)scanf("%d",&mat[i][j]);
}int main(){int n;//freopen("1.out","r",stdin);while(scanf("%d",&n)==1){Init();Input_data(n);printf("min=%d\n",BFS(n));}
}

建图跑迪杰斯特拉：

#include<iostream>
#include<cstring>
#include<cstdio>
#define _cp(a,b)((a.val)<(b.val))using namespace std;const int MAXN = 100000;
const int INF = (1<<30);
const int N = 100 + 5;
int mat[N][N],D[N*N];
bool visit[N*N];
int Size[N*N];
typedef struct node{short val,to;
}Node;
typedef Node elem_t;
Node edge[N*N][4];
const int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
void Init(){for(int i=0;i<N*N;i++){Size[i] = 0;visit[i] = false;D[i] = INF;}
}
void build_mat(int n){Node t;int to,newx,newy,from;for(int i=0;i<n;i++)for(int j=0;j<n;j++){from = i*n+j;for(int d=0;d<4;d++){newx = i + dir[d][0];newy = j + dir[d][1];if(newx>=0 && newx<n && newy>=0 && newy<n && mat[newx][newy]){to = newx*n+newy;int & cnt = Size[from];t.to = to,t.val=mat[newx][newy];edge[from][cnt++] = t;}}}
}
struct heap{elem_t h[MAXN];int n,p,c;void init(){n=0;}void ins(elem_t e){for(p=++n;p>1 && _cp(e,h[p>>1]);h[p]=h[p>>1],p>>=1);h[p] = e;}int del(elem_t &e){if(!n) return 0;for(e=h[p=1],c=2;c<n && _cp(h[c+=(c<n-1 &&_cp(h[c+1],h[c]))],h[n]);h[p] = h[c],p=c,c<<=1);h[p] = h[n--]; return 1;}
};
int DIJ(int n){heap H;H.init();D[0] = 0;Node t;int u,v;t.to = 0,t.val = 0;H.ins(t);while(H.del(t)){u = t.to;visit[u] = true;if(D[u] < t.val) continue;for(int j=0; j<Size[u]; j++){v = edge[u][j].to;if(visit[v]) continue;if(D[v] > D[u] + edge[u][j].val){D[v] = D[u] +edge[u][j].val;t.to = v,t.val = D[v];H.ins(t);}}}return D[n*n-1] + mat[0][0];
}
void Input_data(int n){for(int i=0;i<n;i++)for(int j=0;j<n;j++){scanf("%d",&mat[i][j]);}
}
int main(){int n;while(scanf("%d",&n)==1){Init();Input_data(n);build_mat(n);printf("min=%d\n",DIJ(n));}
}

双向BFS：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>using namespace std;const int N = 100+5;
const int MaxSize = 1e5;
const int INF = (1<<30);
int mat[N][N];
struct visit{int val,vis;
}Min[N][N];typedef struct node{int x,y,val;bool operator < (const node x) const {return val > x.val;
    }
}Node;const int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int BFS(int n){priority_queue<Node> Q1,Q2;int newx,newy,cnt,x=0,y=0,goalx= n-1,goaly=n-1,ans = INF;Node t,s;t.x = x,t.y = y,t.val = mat[0][0];s.x = goalx,s.y = goaly,s.val = mat[goalx][goaly];Min[x][y].vis = 1,Min[x][y].val = mat[x][y];Min[goalx][goaly].vis = 2,Min[goalx][goaly].val = mat[goalx][goaly];Q1.push(t);Q2.push(s);while(!Q1.empty()|| !Q2.empty()){
    cnt = Q1.size();
    while(cnt--){        t = Q1.top(); Q1.pop();
        if(t.x == goalx && t.y == goaly ) ans = min(ans,t.val);         //这里一定要注意，要拓展完才能确定最小值，不能相遇就跳出
        for(int d=0;d<4;d++){            newx = t.x+dir[d][0];
            newy = t.y+dir[d][1];
            if(newx>=0 && newx<n && newy>=0 && newy<n){                if(Min[newx][newy].vis==2) ans = min(t.val + Min[newx][newy].val,ans);
                if(t.val + mat[newx][newy] < Min[newx][newy].val){                s.x = newx,s.y = newy,s.val = t.val+mat[newx][newy];
                Min[newx][newy].val = s.val;
                Min[newx][newy].vis = 1;
                Q1.push(s);
                }
            }
        }
    }cnt = Q2.size();
    while(cnt--){        t = Q2.top();Q2.pop();
        if(t.x == x && t.y == y) return t.val;
        for(int d=0;d<4;d++){            newx = t.x + dir[d][0];
            newy = t.y + dir[d][1];
            if(newx>=0 && newx<n && newy>=0 && newy<n ){                if(Min[newx][newy].vis==1) ans = min(t.val + Min[newx][newy].val,ans);
                if(t.val + mat[newx][newy] < Min[newx][newy].val){                    s.x = newx,s.y = newy,s.val = t.val + mat[newx][newy];
                    Min[newx][newy].val = s.val;
                    Min[newx][newy].vis = 2;
                    Q2.push(s);
                }
            }
        }
    }}return ans;
}void Init(){
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++){        Min[i][j].val = INF;
        Min[i][j].vis = 0;
        }
}void Input_data(int n){
    int val;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++){        scanf("%d",&val);
        mat[i][j] = val?val:INF;
        }
}int main(){
    int n;
    //freopen("1.out","r",stdin);
    while(scanf("%d",&n)==1){    Init();
    Input_data(n);
    printf("min=%d\n",BFS(n));
    }
}
//如有错误,还请指出

在这几种方法里，双向BFS和迪杰斯特拉耗时都是12ms，比其他两种BFS耗时都短，虽然说在这里使用迪杰斯特拉有点大材小用的意味，但是这种方法也去可以应用到其他问题上