• Constructing Roads ——最小生成树


    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

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  • 原文地址:https://www.cnblogs.com/liuxueyang/p/3100811.html
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