Instantaneous Transference--POJ3592Tarjan缩点+搜索

Instantaneous Transference

Time Limit: 5000MS Memory Limit: 65536K

Description

It was long ago when we played the game Red Alert. There is a magic function for the game objects which is called instantaneous transfer. When an object uses this magic function, it will be transferred to the specified point immediately, regardless of how far it is.

Now there is a mining area, and you are driving an ore-miner truck. Your mission is to take the maximum ores in the field.

The ore area is a rectangle region which is composed by n × m small squares, some of the squares have numbers of ores, while some do not. The ores can’t be regenerated after taken.

The starting position of the ore-miner truck is the northwest corner of the field. It must move to the eastern or southern adjacent square, while it can not move to the northern or western adjacent square. And some squares have magic power that can instantaneously transfer the truck to a certain square specified. However, as the captain of the ore-miner truck, you can decide whether to use this magic power or to stay still. One magic power square will never lose its magic power; you can use the magic power whenever you get there.

Input

The first line of the input is an integer T which indicates the number of test cases.

For each of the test case, the first will be two integers N, M (2 ≤ N, M ≤ 40).

The next N lines will describe the map of the mine field. Each of the N lines will be a string that contains M characters. Each character will be an integer X (0 ≤ X ≤ 9) or a ‘’ or a ‘#’. The integer X indicates that square has X units of ores, which your truck could get them all. The ‘’ indicates this square has a magic power which can transfer truck within an instant. The ‘#’ indicates this square is full of rock and the truck can’t move on this square. You can assume that the starting position of the truck will never be a ‘#’ square.

As the map indicates, there are K ‘’ on the map. Then there follows K lines after the map. The next K lines describe the specified target coordinates for the squares with ‘‘, in the order from north to south then west to east. (the original point is the northwest corner, the coordinate is formatted as north-south, west-east, all from 0 to N - 1,M - 1).

Output

For each test case output the maximum units of ores you can take. 　

Sample Input

1
2 2
11
1*
0 0

Sample Output

3

Source

South Central China 2008 hosted by NUDT

题意：在一个矿区，你驾驶着一辆采矿的卡车，你的任务是采集最大数量的矿石，矿区是一个长方形的区域，包含n*m个的小方块，有些矿区有矿石，有的没有，矿石采完后不能再生，采矿车的起始位置在西北角（0,0），它只能向东面或者南面相邻的方格，其中的一些方块有魔法，能将矿车瞬间移动到指定的位置，作为驾驶员，你可以决定是否使用这个魔法，魔法不会消失。

思路：首先是建图，建完图发现会形成强连通分量，所以我们先进行缩点，缩完点后会形成一个DAG图，然后进行DFS搜索一边就可以，不过在建图的时候要注意传送到外面的点也要建图

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <string>#include <queue>#include <stack>#include <set>#include <algorithm>using namespace std;const int Max = 2000;typedef struct node{    int v,next;}Line;Line Li[Max*1000];int Head1[Max],Head2[Max],top;int dfn[Max],low[Max],vis[Max],dep,pre[Max];int Dp[Max],num;char str[50][50];int va[Max],a[Max];int T,n,m;int dir[][2]={{0,1},{1,0}};stack<int>S;void Init(){    memset(Head1,-1,sizeof(Head1));    memset(Head2,-1,sizeof(Head2));    memset(vis,0,sizeof(vis));    memset(va,0,sizeof(va));    memset(a,0,sizeof(a));    memset(pre,-1,sizeof(pre));    dep = 0 ;num =  0;}void AddEdge1(int u,int v){    Li[top].v = v; Li[top].next =Head1[u];    Head1[u] = top++;}void AddEdge2(int u,int v){    Li[top].v = v; Li[top].next = Head2[u];    Head2[u] = top++;}bool Judge(int x,int y){    if(x>=0&&x<n&&y>=0&&y<m&&str[x][y]!='#')    {        return true;    }    return false;}void DFS(int x,int y){    for(int i = 0;i<2;i++)    {        int Fx = x+dir[i][0];        int Fy = y+dir[i][1];        if(Judge(Fx,Fy))        {            AddEdge1(x*m+y,Fx*m+Fy);        }    }}void Tarjan(int u) //Tarjan强连通缩点{    dfn[u] = low[u] =dep++;    vis[u]=1;    S.push(u);    for(int i=Head1[u];i!=-1;i=Li[i].next)    {        if(vis[Li[i].v]==1)        {            low[u] = min(low[u],dfn[Li[i].v]);        }        else if(vis[Li[i].v]==0)        {            Tarjan(Li[i].v);            low[u] = min(low[u],low[Li[i].v]);        }    }    if(dfn[u]==low[u])    {        while(!S.empty())        {            int v=S.top();            S.pop();            pre[v] = num;            vis[v] = 2;            a[num]+=va[v];            if(u==v)            {                break;            }        }        num++;    }}int dfs(int u)//搜索最大值{    if(!vis[u])    {        vis[u]=1;        int ans=0;        for(int i=Head2[u];i!=-1;i=Li[i].next)        {            ans = max(ans,dfs(Li[i].v));        }        a[u]+=ans;    }    return a[u];}int main(){    scanf("%d",&T);    int x,y;    while(T--)    {        scanf("%d %d",&n,&m);        Init();        for(int i=0;i<n;i++)        {            scanf("%s",str[i]);        }        for(int i=0;i<n;i++)//建图        {            for(int j=0;j<m;j++)            {                if(str[i][j] == '#')                {                    continue;                }                DFS(i,j);                if(str[i][j]=='*')                {                    scanf("%d %d",&x,&y);                    AddEdge1(i*m+j,x*m+y);                }                else                {                    va[i*m+j]=str[i][j]-'0';                }            }        }        for(int i=0;i<n*m;i++)        {            if(!vis[i])            {                Tarjan(i);            }        }        for(int i=0;i<n*m;i++) //重新建图        {            for(int j=Head1[i];j!=-1;j = Li[j].next)            {                if(pre[i]!=pre[Li[j].v])                {                    AddEdge2(pre[i],pre[Li[j].v]);                }            }        }        memset(vis,0,sizeof(vis));        printf("%d\n",dfs(pre[0]));    }    return 0;}/*110 1011678116781*778116781*700016781*778116781#778000781#77837###1*00037###1*34000###1*3451*77837###1#3455 55 55 65 75 85 95 10ans=130*/