http://poj.org/problem?id=3592
/*
题意：给定一个n*m的矩阵，你从左上角出发，规定只能往当前点的右边或者下边走，其中还有一些特殊点*具有特殊的力量可以把你传到特定的一个点（你可以选择传送也可以选择不传送），问从左上角出发到不能走下去，最多能获得的矿石量（每个方格对应着一个数字表示矿石数量）。点#直接跳过
思路：首先build1根据题意描述，见图，将二位矩阵转化为一维的点建图，每个点可以向右向下建立有向边，点*还可以向传送点建边。建完后tarjan缩点，然后在根据缩点后的点建图，添加超级源点s，权值为i-j sum[j], 求最长路即可的结果;
中间数组开成了44贡献了几次wa，转化为1维后是40*40了。。。
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#define maxn 44
#define N 2002
using namespace std;int dir[2][2] = {{0,1},{1,0}};
const int inf = 99999999;struct node
{int v,w;int next;
}g1[N*N],g2[N*N];int head1[N],head2[N],ct1,ct2;
int low[N],dfn[N],stack[N],val[N];
int bcnt,top,index,num[maxn][maxn],kpos[N];
int belong[N],sum[N],dis[N];
bool inq[N],ins[N];
int n,m,ct,knum,s;
char str[maxn][maxn];void add1(int u,int v)
{g1[ct1].v = v;g1[ct1].next = head1[u];head1[u] = ct1++;
}
void add2(int u,int v,int w)
{g2[ct2].v = v;g2[ct2].w =w;g2[ct2].next = head2[u];head2[u] = ct2++;
}
void tarjan(int i)
{int k,j;low[i] = dfn[i] = ++index;ins[i] = true;stack[++top] = i;for (k = head1[i]; k != -1; k = g1[k].next){int j = g1[k].v;if (!dfn[j]){tarjan(j);low[i] = min(low[i],low[j]);}else if (ins[j]){low[i] = min(low[i],dfn[j]);}}if (dfn[i] == low[i]){bcnt++;do{j = stack[top--];ins[j] = false;belong[j] = bcnt;}while (j != i);}
}
void build1()
{int i,j,k,x,y;memset(num,0,sizeof(num));memset(kpos,0,sizeof(kpos));memset(val,0,sizeof(val));ct = knum = 0;scanf("%d%d",&n,&m);for (i = 0; i < n; ++i){scanf("%s",str[i]);for (j = 0; j < m; ++j){if (str[i][j] == '#') continue;num[i][j] = ++ct;if (str[i][j] >= '0' && str[i][j] <= '9')val[num[i][j]] = str[i][j] - '0';else{kpos[knum++] = num[i][j];//记录每个*点，应为后边依次输入其传输的坐标val[num[i][j]] = 0;}}}memset(head1,-1,sizeof(head1));ct1 = 0;for (i = 0; i < n; ++i){for (j = 0; j < m; ++j){if (str[i][j] == '#') continue;for (k = 0; k < 2; ++k){int tx = i + dir[k][0];int ty = j + dir[k][1];if (tx >= 0 && tx < n && ty >= 0 && ty < m)add1(num[i][j],num[tx][ty]);}}}//根据*点传输的坐标添加边for (i = 0; i < knum; ++i){scanf("%d%d",&x,&y);add1(kpos[i],num[x][y]);}memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));memset(ins,false,sizeof(ins));memset(belong,0,sizeof(belong));top = index = bcnt = 0;for (i = 1; i <= ct; ++i){if (!dfn[i]) tarjan(i);//tarjan缩点}//缩点后计算权值memset(sum,0,sizeof(sum));for (i = 1; i <= ct; ++i)sum[belong[i]] += val[i];
}
void build2()
{int i,k;s = 0;memset(head2,-1,sizeof(head2));//加入超级源点ct2 = 0;add2(s,belong[1],sum[belong[1]]);for (i = 1; i <= ct; ++i){for (k = head1[i]; k != -1; k = g1[k].next){int j = g1[k].v;if (belong[i] != belong[j]){add2(belong[i],belong[j],sum[belong[j]]);}}}
}
void spfa(int s)
{int i;queue<int>q;for (i = 0; i < N; ++i){dis[i] = -inf;inq[i] = false;}q.push(s); dis[s] = 0;inq[s] = true;while (!q.empty()){int u = q.front(); q.pop();inq[u] = false;for (i = head2[u]; i != -1; i = g2[i].next){int v = g2[i].v;if (dis[v] < dis[u] + g2[i].w){dis[v] = dis[u] + g2[i].w;if (!inq[v]){inq[v] = true;q.push(v);}}}}
}
void solve()
{spfa(s);//求最长路int ans = 0;for (int i = 0; i <= bcnt; ++i)ans = max(ans,dis[i]);printf("%d\n",ans);
}
int main()
{int t;scanf("%d",&t);while (t--){build1();//根据题意建图build2();//缩点后建图solve();//求解}return 0;
}