pku 3592 Instantaneous Transference tarjan缩点重建图+spfa求最长路
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;
}