POJ1548 Robots "Minimum path overlay of the binary graph"

Topic Links:

http://poj.org/problem?id=1548

Main topic:

in a n*m (N <= 24,m <= 24) Figure, there is a lot of rubbish, and the robot that cleans the rubbish starts from the upper left corner. have been

know that the robot can only clean the rubbish to the right or down, after cleaning up a place to continue to the right or down to clear Management

Other rubbish. End-of-run to (n,m) position terminated. Q: The minimum number of robots required to clean up all the rubbish.

Ideas:

The figure does not give the size of N and M, just gives the location of the rubbish. The input ends with 0 0 for a set of data inputs. Build a structure

To store the garbage's coordinate values. Now to build a binary map, the two sides of the graph is the garbage node, traverse the original, if Garbage J in

in the lower right corner of the garbage I, the (i,j) is added to the binary diagram, and the problem becomes: the minimum path coverage of the binary graph is obtained. by two

Min -Path overlay = Point-two graph maximum matching, according to the Hungarian algorithm to calculate the maximum matching of two, and then to obtain the knot The results.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace    std;const int MAXN = 600;struct node{int x; int y;} Point[maxn];bool map[maxn][maxn],mask[maxn];int nx,ny;int cx[maxn],cy[maxn];int FindPath (int u) {for (int i = 0; i < NY; ++i) {if (Map[u][i] &&!            Mask[i]) {mask[i] = 1; if (cy[i] = =-1 | |                Findpath (Cy[i])) {Cy[i] = u;                Cx[u] = i;            return 1; }}} return 0;}    int Maxmatch () {for (int i = 0; i < NX; ++i) cx[i] = 1;    for (int i = 0; i < NY; ++i) cy[i] = 1;    int res = 0;                for (int i = 0, i < NX; ++i) {if (cx[i] = = 1) {for (int j = 0; j < NY; ++j)            MASK[J] = 0;        Res + = Findpath (i); }} return res;}    int main () {int id = 0; while (scanf ("%d%d", &point[id].x,&point[id].y) && (point[id].x!=-1 && point[id].y!=-1)) {id++; while (scanf ("%d%d", &point[id].x,&point[id].y) && (point[id].x| |        POINT[ID].Y)) id++;        memset (map,0,sizeof (MAP)); for (int i = 0, i < ID; ++i) {for (int j = i+1; j < ID; ++j) {if (point[i                ].x <= point[j].x && point[i].y <= point[j].y) map[i][j] = 1;            else if (point[i].x >= point[j].x && point[i].y >= point[j].y) map[j][i] = 1;        }} NX = NY = ID;        printf ("%d\n", Id-maxmatch ());    id = 0; } return 0;}

POJ1548 Robots "Minimum path overlay of the binary graph"