Aizu 2302 On or Off

传送门
暴力模拟
因为每个房间之间只有一条路，所以不用什么决策或者dp
dfs出每个点经过的时间，然后比较，cost[i][j]×(Tnow−Tpre)与on[i][j]+off[i][j]哪个小就选哪个。
第一次经过一定要开灯和最后一次经过一定要关灯。所以只要经过了就需要加上on[i][j]+off[i][j]。

//      whn6325689//      Mr.Phoebe//      http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<50#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,r#define root 1,1,ntemplate<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------int ma[55][55],r,c,m;vector<int> t[55][55];pii road[3333];int on[55][55],off[55][55],cost[55][55];const int dir[4][2]= {0,1,0,-1,1,0,-1,0};char str[55];bool vis[55][55];pii task[1111];inline bool in(int x,int y){    return x>=0 && x<r && y>=0 && y<c;}int dfs(int x,int y,int ex,int ey,int step){    if(x==ex && y==ey)    {        return step;    }    for(int i=0; i<4; i++)    {        int xx=x+dir[i][0];        int yy=y+dir[i][1];        if(in(xx,yy) && ma[xx][yy] && !vis[xx][yy])        {            road[step]=mp(xx,yy);            vis[xx][yy]=1;            int temp=dfs(xx,yy,ex,ey,step+1);            if(temp!=-1)                return temp;            vis[xx][yy]=0;        }    }    return -1;}int main(){//freopen("data.txt","r",stdin);    while(scanf("%d %d %d",&r,&c,&m)!=EOF)    {        for(int i=0; i<r; i++)        {            scanf("%s",str);            for(int j=0; j<c; j++)            {                t[i][j].clear();                if(str[j]=='#')                    ma[i][j]=0;                else                    ma[i][j]=1;            }        }        for(int i=0; i<r; i++)            for(int j=0; j<c; j++)                read(cost[i][j]);        for(int i=0; i<r; i++)            for(int j=0; j<c; j++)                read(on[i][j]);        for(int i=0; i<r; i++)            for(int j=0; j<c; j++)                read(off[i][j]);        for(int i=0; i<m; i++)            read(task[i].first),read(task[i].second);        t[task[0].first][task[0].second].pb(0);        int time=0;        for(int i=1; i<m; i++)        {            CLR(vis,0);            vis[task[i-1].first][task[i-1].second]=1;            int step=dfs(task[i-1].first,task[i-1].second,task[i].first,task[i].second,0);            for(int j=0; j<step; j++)                t[road[j].first][road[j].second].push_back(time+j+1);            time+=step;        }        int ans=0;        for(int i=0; i<r; i++)            for(int j=0; j<c; j++)                for(int k=0; k<t[i][j].size(); k++)                {                    if(k)                    {                        if((t[i][j][k]-t[i][j][k-1])*cost[i][j] < on[i][j]+off[i][j])                            ans+=(t[i][j][k]-t[i][j][k-1])*cost[i][j];                        else                            ans+=on[i][j]+off[i][j];                    }                    else                        ans+=on[i][j]+off[i][j];                }        write(ans),putchar('\n');    }    return 0;}