Instantaneous Transference
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 6177 | Accepted: 1383 |
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
题目大意：
有一个N*M的矩阵地图，矩阵中用了多种字符代表不同的地形，如果是数字X(0~9)，则表示该区域为矿区，有X单位的矿产。如果是"*"，则表示该区域为传送点，并且对应唯一一个目标坐标。如果是"#",，则表示该区域为山区，矿车不能进入。现在矿车的出发点在坐标(0，0)点。并且(0，0)点一定不是"#"区域。矿车只能向右走、向下走或是遇到传送点的时候可以传送到指定位置。那么问题来了：矿车最多能采到多少矿。
思路：
如果把N*M个矩阵单位看做是N*M个点，编号为0~N*M。然后从一个坐标到另一个坐标看做是两点之间的边。到达的坐标所拥有的矿产为边的权值。那么问题就变成了：矿车从节点0出发，所能达到的最长路径。但是除了向右走和向下走的边，考虑到还有传送点和目标坐标构成的边，原图上就会多了很多回退边，构成了很多的有向环。有向环的出现，使得矿车能够采到的矿产增多了一部分，只要能走到有向环内，则该环内所有点的矿产都能被采到。但是问题也出来了，如果不做处理，直接搜索路径，那么矿车很可能会走进环内不出来。于是想到了缩点，把有向环缩为一个点，也就是强连通分量缩点。并记录强连通分量中的总矿产值。缩点后，原图就变成了一个有向无环图(DAG)。然后重新建立一个新图(DAG)，对新图求最长路径(用SPFA算法)，得到源点(0，0)到各点的最长路径。从中找出最长的路径，就是所求的结果。
这题和POJ3126类似，都是缩点SPFA求最长路，POJ312解析
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#define maxn 2000+100
#define maxm 40000+100
#define INF 0x3f3f3f3f
using namespace std;
int n, m;
struct node {
int u, v, next;
};
node edge[maxm];
int head[maxn], cnt;
int low[maxn], dfn[maxn];
int dfs_clock;
int Stack[maxn], top;
bool Instack[maxn];
int Belong[maxn];
int scc_clock;
int val[maxn];//存每个点的矿石量
int sumval[maxn];//存每个缩点的矿石量
vector<int>Map[maxm];
char map[100][100];
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
memset(val, 0, sizeof(val));
memset(sumval, 0, sizeof(sumval));
memset(val, 0, sizeof(val));
}
void addedge(int u, int v){
edge[cnt] = {u, v, head[u]};
head[u] = cnt++;
}
void getmap(){
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%s", map[i]);
for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j){
if(map[i][j] == '#') continue;
if(i + 1 < n && map[i + 1][j] != '#')//向下走
addedge(i * m + j, (i + 1) * m + j);
if(j + 1 < m && map[i][j + 1] != '#')//向右走
addedge(i * m + j, i * m + j + 1);
val[i * m + j] = map[i][j] - '0';
if(map[i][j] == '*'){
val[i * m + j] = 0;
int x, y;
scanf("%d%d", &x, &y);
if(map[x][y] != '#');//传送的位置可能为 #
addedge(i * m + j, x * m + y);
}
}
}
}
void Tarjan(int u){
int v;
low[u] = dfn[u] = ++dfs_clock;
Stack[top++] = u;
Instack[u] = true;
for(int i = head[u]; i != -1; i = edge[i].next){
int v = edge[i].v;
if(!dfn[v]){
Tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(Instack[v])
low[u] = min(low[u], dfn[v]);
}
if(dfn[u] == low[u]){
scc_clock++;
do{
v = Stack[--top];
sumval[scc_clock] += val[v];
Instack[v] = false;
Belong[v] = scc_clock;
}
while( v != u);
}
}
void find(){
memset(low, 0, sizeof(low));
memset(dfn, 0, sizeof(dfn));
memset(Belong, 0, sizeof(Belong));
memset(Stack, 0, sizeof(Stack));
memset(Instack, false, sizeof(false));
dfs_clock = scc_clock = top = 0;
for(int i = 0; i < n * m; ++i){
if(!dfn[i])
Tarjan(i);
}
}
void suodian(){//缩点新建图
for(int i = 1; i <= scc_clock; ++i)
Map[i].clear();
// for(int i = 0; i < n * m; ++i){
// for(int j = head[i]; j != -1; j = edge[j].next){
// int u = Belong[i];
// int v = Belong[edge[j].v];
// if(u != v)
// Map[u].push_back(v);
// }
// }
//上面也是一种建图方式。
for(int i = 0; i < cnt; ++i){
int u = Belong[edge[i].u];
int v = Belong[edge[i].v];
if(u != v)
Map[u].push_back(v);
}
}
int vis[maxn],dist[maxn];
void SPFA(){
queue<int>q;
memset(vis, 0, sizeof(vis));
memset(dist, 0, sizeof(dist));
vis[Belong[0]] = 1;
dist[Belong[0]] = sumval[Belong[0]];
q.push(Belong[0]);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = 0;
for(int i = 0; i < Map[u].size(); ++i){
int v = Map[u][i];
if(dist[v] < dist[u] + sumval[v]){
dist[v] = dist[u] + sumval[v];
if(!vis[v]){
vis[v] = 1;
q.push(v);
}
}
}
}
}
int main (){
int T;
scanf("%d", &T);
while(T--){
init();
getmap();
find();
suodian();
SPFA();
sort(dist + 1, dist + scc_clock + 1);
printf("%d\n", dist[scc_clock]);
}
return 0;
}
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