sicily1321. Robot
1321. Robot
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
Description
Karell Incorporated has designed a new exploration robot that has the ability to explore new terrains, this new robot can move in all kinds of terrain, it only needs more fuel to move in rough terrains, and less fuel in plain terrains. The only problem this robot has is that it can only move orthogonally, the robot can only move to the grids that are at the North, East, South or West of its position.
The Karell`s robot can communicate to a satellite dish to have a picture of the terrain that is going to explore, so it can select the best route to the ending point, The robot always choose the path that needs the minimum fuel to complete its exploration, however the scientist that are experimenting with the robot, need a program that computes the path that would need the minimum amount of fuel. The maximum amount of fuel that the robot can handle is 9999 units
The Terrain that the robot receives from the satellite is divided into a grid, where each cell of the grid is assigned to the amount of fuel the robot would need to pass thought that cell. The robot also receives the starting and ending coordinates of the exploration area.
Path Example
From (1,1) to (5,5)
Fuel needed 10
Input
The first line of the input file is the number of tests that must be examined.
The first line of the test is the number of rows and columns that the grid will contain. The rows and columns will be 0 < row100, 0 < column100
The next lines are the data of the terrain grid
The last line of the test has the starting and ending coordinates.
Output
One line, for each test will have the amount of fuel needed by the robot
Sample Input
Sample Output
この問題図はマトリクスに格納されており,一度遍歴すると最短経路を見つけることができる.
タイトルリンク:http://soj.me/1321
参照コード:
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
Description
Karell Incorporated has designed a new exploration robot that has the ability to explore new terrains, this new robot can move in all kinds of terrain, it only needs more fuel to move in rough terrains, and less fuel in plain terrains. The only problem this robot has is that it can only move orthogonally, the robot can only move to the grids that are at the North, East, South or West of its position.
The Karell`s robot can communicate to a satellite dish to have a picture of the terrain that is going to explore, so it can select the best route to the ending point, The robot always choose the path that needs the minimum fuel to complete its exploration, however the scientist that are experimenting with the robot, need a program that computes the path that would need the minimum amount of fuel. The maximum amount of fuel that the robot can handle is 9999 units
The Terrain that the robot receives from the satellite is divided into a grid, where each cell of the grid is assigned to the amount of fuel the robot would need to pass thought that cell. The robot also receives the starting and ending coordinates of the exploration area.
Path Example
From (1,1) to (5,5)
Fuel needed 10
Input
The first line of the input file is the number of tests that must be examined.
The first line of the test is the number of rows and columns that the grid will contain. The rows and columns will be 0 < row100
The next lines are the data of the terrain grid
The last line of the test has the starting and ending coordinates.
Output
One line, for each test will have the amount of fuel needed by the robot
Sample Input
3
5 5
1 1 5 3 2
4 1 4 2 6
3 1 1 3 3
5 2 3 1 2
2 1 1 1 1
1 1 5 5
5 4
2 2 15 1
5 1 15 1
5 3 10 1
5 2 1 1
8 13 2 15
1 1 1 4
10 10
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 10 10
Sample Output
10
。 ， 。 この問題図はマトリクスに格納されており,一度遍歴すると最短経路を見つけることができる.
タイトルリンク:http://soj.me/1321
参照コード:
#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
using namespace std;
int dis[109][109];
struct Gra
{
int r;
int c;
friend bool operator <(Gra g1, Gra g2)
{
if(dis[g1.r][g1.c] < dis[g2.r][g2.c])
return true;
}
};
int main()
{
int t;
cin >> t;
while (t --)
{
int a, b;
scanf("%d%d", &a, &b);
int s[109][109];
for (int i = 1; i <= a; ++ i)
{
for (int j = 1; j <= b; ++ j)
{
cin >> s[i][j];
}
}
int ar, ac, br, bc;
cin >> ar >> ac >> br >> bc;
set<Gra> q;
int know[109][109];
memset(know, 0, sizeof(know));
memset(dis, 9999999, sizeof(dis));
Gra tu[10009];
tu[1].r = ar;
tu[1].c = ac;
q.insert(tu[1]);
Gra temp;
int i = 2;
//know[ar][ac] = 1;
dis[ar][ac] = s[ar][ac];
set<Gra>::iterator it;
set<Gra>::iterator itt;
while (!q.empty())
{
it = q.begin();
temp = *it;
itt = it;
++ it;
Gra tem;
for (; it != q.end(); ++ it)
{
tem = *it;
if (dis[temp.r][temp.c] > dis[tem.r][tem.c])
{
temp = tem;
itt = it;
}
}
q.erase(itt);
know[temp.r][temp.c] = 1;
if (temp.r == br && temp.c == bc)
{
cout << dis[temp.r][temp.c] << endl;
break;
}
if (temp.c - 1 >= 1)
{
tu[i].c = temp.c - 1;
tu[i].r = temp.r;
if (know[tu[i].r][tu[i].c] == 0)
{
if (dis[tu[i].r][tu[i].c] > dis[temp.r][temp.c] + s[tu[i].r][tu[i].c])
{
dis[tu[i].r][tu[i].c] = dis[temp.r][temp.c] + s[tu[i].r][tu[i].c];
q.insert(tu[i]);
++ i;
}
}
}
if (temp.r - 1 >= 1)
{
tu[i].r = temp.r - 1;
tu[i].c = temp.c;
if (know[tu[i].r][tu[i].c] == 0)
{
if (dis[tu[i].r][tu[i].c] > dis[temp.r][temp.c] + s[tu[i].r][tu[i].c])
{
dis[tu[i].r][tu[i].c] = dis[temp.r][temp.c] + s[tu[i].r][tu[i].c];
q.insert(tu[i]);
++ i;
}
}
}
if (temp.c + 1 <= b)
{
tu[i].c = temp.c + 1;
tu[i].r = temp.r;
if (know[tu[i].r][tu[i].c] == 0)
{
if (dis[tu[i].r][tu[i].c] > dis[temp.r][temp.c] + s[tu[i].r][tu[i].c])
{
dis[tu[i].r][tu[i].c] = dis[temp.r][temp.c] + s[tu[i].r][tu[i].c];
q.insert(tu[i]);
++ i;
}
}
}
if (temp.r + 1 <= a)
{
tu[i].r = temp.r + 1;
tu[i].c = temp.c;
if (know[tu[i].r][tu[i].c] == 0)
{
if (dis[tu[i].r][tu[i].c] > dis[temp.r][temp.c] + s[tu[i].r][tu[i].c])
{
dis[tu[i].r][tu[i].c] = dis[temp.r][temp.c] + s[tu[i].r][tu[i].c];
q.insert(tu[i]);
++ i;
}
}
}
}
}
//system("pause");
}