URAL 2005. Taxi for Programmers

2005. Taxi for Programmers

Time limit: 0.5 second
Memory limit: 64 MB

The clock shows 11:30 PM. The sports programmers of the institute of maths and computer science have just finished their training. The exhausted students gloomily leave their computers. But there’s something that cheers them up: Misha, the kind coach is ready to give all of them a lift home in his brand new car. Fortunately, there are no traffic jams on the road. Unfortunately, all students live in different districts. As Misha is a programmer, he highlighted and indexed five key points on the map of Yekaterinburg: the practice room (1), Kirill’s home (2), Ilya’s home (3), Pasha’s and Oleg’s home (4; they live close to each other) and his own home (5). Now he has to find the optimal path. The path should start at point 1, end at point 5 and have minimum length. Moreover, Ilya doesn’t want to be the last student to get home, so point 3 can’t be fourth in the path.

Input

The input contains a table of distances between the key points. It has five rows and five columns. The number in the i’th row and the j’th column (1 ≤ i, j ≤ 5) is a distance between points i and j. All distances are non-negative integers not exceeding 10 000. It is guaranteed that the distance from any point to itself equals 0, and for any two points, the distance between them is the same in both directions. It is also guaranteed that the distance between any two points doesn’t exceed the total distance between them through another point.

Output

Output two lines. The first line should contain the length of the optimal path. The second line should contain five space-separated integers that are the numbers of the points in the order Misha should visit them. If the problem has several optimal solutions, you may output any of them.

Sample

inputoutput

0 2600 3800 2600 25002600 0 5300 3900 44003800 5300 0 1900 45002600 3900 1900 0 37002500 4400 4500 3700 0

135001 2 3 4 5

题意：有限制的最短路。

解析：虽然限制3号点不能第四个到达，但是顶点数固定是5个，这样其实就没必要用最短路算法了，直接预处理出最短路径的所有可能，然后选择最短的那条就行了。

AC代码：

#include <cstdio>#include <algorithm>using namespace std;const int n = 5;int a[6][6];int act[][5] = {{1, 2, 3, 4, 5}, {1, 3, 2, 4, 5}, {1, 3, 4, 2, 5}, {1, 4, 3, 2, 5}};      //列举出所有符合条件的路径int main(){    #ifdef sxk        freopen("in.txt", "r", stdin);    #endif //sxk    for(int i=1; i<=n; i++)        for(int j=1; j<=n ;j++) scanf("%d", &a[i][j]);    int foo = 0, ans = 123456789;    for(int i=0; i<4; i++){        int sum = 0;        for(int j=0; j<n-1; j++){            sum += a[ act[i][j] ][ act[i][j+1] ];        }        if(ans > sum){            ans = sum;            foo = i;        }    }    printf("%d\n", ans);    for(int i=0; i<n; i++) printf("%d%c", act[foo][i], i == n-1 ? '\n' : ' ');    return 0;}