HDU 3152 Obstacle Course(优先队列)

[align=left]Problem Description[/align]



You are working on the team assisting with programming for the Mars rover. To conserve energy, the rover needs to find optimal paths across the rugged terrain to get from its starting location to its final location. The following is the first approximation
for the problem.



N * N square matrices contain the expenses for traversing each individual cell. For each of them, your task is to find the minimum-cost traversal from the top left cell [0][0] to the bottom right cell [N-1][N-1].
Legal moves are up, down, left, and right; that is, either the row index changes by one or the column index changes by one, but not both.

 

[align=left]Input[/align]
Each problem is specified by a single integer between 2 and 125 giving the number of rows and columns in the
N * N square matrix. The file is terminated by the case
N = 0.

Following the specification of N you will find N lines, each containing
N numbers. These numbers will be given as single digits, zero through nine, separated by single blanks.

 

[align=left]Output[/align]
Each problem set will be numbered (beginning at one) and will generate a single line giving the problem set and the expense of the minimum-cost path from the top left to the bottom right corner, exactly as shown in the sample output
(with only a single space after "Problem" and after the colon).

 

[align=left]Sample Input[/align]

3
5 5 4
3 9 1
3 2 7
5
3 7 2 0 1
2 8 0 9 1
1 2 1 8 1
9 8 9 2 0
3 6 5 1 5
7
9 0 5 1 1 5 3
4 1 2 1 6 5 3
0 7 6 1 6 8 5
1 1 7 8 3 2 3
9 4 0 7 6 4 1
5 8 3 2 4 8 3
7 4 8 4 8 3 4
0

 

[align=left]Sample Output[/align]

Problem 1: 20
Problem 2: 19
Problem 3: 36

 

[align=left]Source[/align]
2008 ACM-ICPC Pacific Northwest Region

 
 
给你一副N*N的地图。
从左上角走到右下角，所经路程的最小花费。（数字即是花费）
因为不是求最短路径，而是求最少的花费。
可以用优先队列，花费小的优先级高。加上BFS，开个flag数字记录任意点的花费。
每次不断的更新flag数组的值，因为优先队列，每次弹出来的必然是优先级最高的（花费少的）
上代码

#include <stdio.h>
#include<queue>
#include <string.h>
#include <algorithm>
using namespace std;
int n;
int map[130][130];
int flag[130][130];   //记录到任意点的花费。
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
#define inf 0x6fffff
struct node
{
int x,y;
int cost;
friend bool operator<(node a,node b)
{
return a.cost>b.cost;  //优先队列，定义优先级，花费小的优先。
}
}
;
bool check(int x,int y)
{
if(x>=n ||y>=n ||x<0||y<0 )
return 0;
return 1;
}
int bfs(int x,int y)
{
int i;
node st,ed;
priority_queue<node>q;  //优先队列
st.x=x;
st.y=y;
st.cost=map[0][0];
memset(flag,-1,sizeof(flag));
q.push(st);
while(!q.empty())
{
st=q.top();  //每次出来的都是最小的花费。
q.pop();
for(i=0;i<4;i++)
{
ed.x=st.x+dir[i][0];
ed.y=st.y+dir[i][1];
if(!check(ed.x,ed.y))//排除越界就够了，不用排除已经访问的，可以走回头路的，反正要遍历整个地图。
continue;
ed.cost=st.cost+map[ed.x][ed.y];
if(ed.cost<flag[ed.x][ed.y]||flag[ed.x][ed.y]==-1)
{
flag[ed.x][ed.y]=ed.cost;  //不断的更新最小花费
if(ed.x!=n-1 ||ed.y!=n-1)
q.push(ed);
}
}
}
return flag[n-1][n-1];  //到达右下角的状态自然为最小的花费
}
int main()
{
int i,j;
int c=0;
while(scanf("%d",&n)&&n)
{
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&map[i][j]);  //输入地图
int ans=bfs(0,0);   //跑一遍BFS即可
printf("Problem %d: %d\n",++c,ans);
}
return 0;
}