【二分图+简单题】杭电 hdu 1151 Air Raid

/* THE PROGRAM IS MADE BY PYY */ /*----------------------------------------------------------------------------// Copyright (c) 2011 panyanyany All rights reserved. URL : http://acm.hdu.edu.cn/showproblem.php?pid=1151 Name : 1151 Air Raid Date : Monday, November 7, 2011 Time Stage : half an hour Result: 4923333 2011-11-07 20:58:18 Accepted 1151 15MS 316K 1557 B C++ pyy Test Data : Review : 注意是有向图 //----------------------------------------------------------------------------*/ #include <stdio.h> #include <vector> using std::vector ; #define MAXSIZE 129 int tcase, iStreet, iIntersection ; int link[MAXSIZE], cover[MAXSIZE] ; int map[MAXSIZE][MAXSIZE] ; vector<int> adj[MAXSIZE] ; int find (int cur) { int i, j ; for (i = 1 ; i <= iIntersection ; ++i) { if (cover[i] == false && map[cur][i]) { cover[i] = true ; if (link[i] == 0 || find (link[i])) { link[i] = cur ; return 1 ; } } } return 0 ; } int main () { int i, j ; int s, e ; int sum, sumLink ; while (~scanf ("%d", &tcase)) { while (tcase--) { scanf ("%d%d", &iIntersection, &iStreet) ; memset (map, 0, sizeof (map)) ; for (i = 1 ; i <= iStreet ; ++i) { scanf ("%d%d", &s, &e) ; map[s][e] = 1 ; // map[e][s] = 1 ; 若加上这一句，则此图变为无向图 } sum = 0 ; memset (link, 0, sizeof (link)) ; for (i = 1 ; i <= iIntersection ; ++i) { memset (cover, 0, sizeof (cover)) ; sum += find (i) ; } sumLink = 0 ; for (i = 1 ; i <= iIntersection ; ++i) { sumLink += link[i] ? 1 : 0 ; } // 由算法的特点可知，若 S, E 两点连通，则 link[S] = 0, link[E] = S ; 因此后面还要减个 sum // 当然，直接 iIntersection – sumLink 也是一样滴～～ printf ("%d\n", sum + (iIntersection - sumLink - sum)) ; } } return 0 ; }