题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=24534#problem/B
题目大意：
给邻接矩阵，和已经建立好的几条边。求最小生成树权值。
题目思路：
方法就是把已将建立好的边的权值赋值为0即可。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 #define MAXN 110 8 #define MAXM 5009 9 typedef struct edge { 10 int u, v, w; 11 bool operator < (const edge &other) const { 12 return w < other.w; 13 } 14 }edge; 15 edge edges[MAXM]; 16 int parent[MAXN]; 17 int n, m, i, j; 18 void init() { 19 for (i = 1; i <= n; ++i) parent[i] = -1; 20 } 21 int find(int x) { 22 int s; 23 for (s = x; parent[s] >= 0; s = parent[s]) ; 24 while (s != x) { 25 int tmp = parent[x]; 26 parent[x] = s; 27 x = tmp; 28 } 29 return s; 30 } 31 void Union(int R1, int R2) { 32 int r1 = find(R1), r2 = find(R2), tmp = parent[r1] + parent[r2]; 33 if (parent[r1] > parent[r2]) { 34 parent[r1] = r2; parent[r2] = tmp; 35 } else { 36 parent[r2] = r1; parent[r1] = tmp; 37 } 38 } 39 void kruscal() { 40 int sum = 0, num = 0, u, v; 41 init(); 42 for (i = 0; i < m; ++i) { 43 u = edges[i].u; v = edges[i].v; 44 if (find(u) != find(v)) { 45 sum += edges[i].w; num++; 46 Union(u, v); 47 } 48 if (num >= n-1) break; 49 } 50 printf("%d\n", sum); 51 } 52 int ma[MAXN][MAXN]; 53 int main(void) { 54 #ifndef ONLINE_JUDGE 55 freopen("hust_b.in", "r", stdin); 56 #endif 57 while (~scanf("%d", &n)) { 58 for (i = 1; i <= n; ++i) { 59 for (j = 1; j <= n; ++j) { 60 scanf("%d", &ma[i][j]); 61 } 62 } 63 m = 0; 64 int q, u, v; scanf("%d", &q); 65 while (q--) { 66 scanf("%d%d", &u, &v); 67 ma[u][v] = ma[v][u] = 0; 68 } 69 for (i = 1; i <= n; ++i) { 70 for (j = 1; j < i; ++j) { 71 edges[m].u = i; edges[m].v = j; edges[m].w = ma[i][j]; 72 m++; 73 } 74 } 75 sort(edges, edges+m); 76 kruscal(); 77 } 78 79 return 0; 80 }
这题以前做过……模板题