题目链接：POJ 2421 Constructing Roads 修建道路
Constructing Roads
| Time Limit: 2000MS |
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Memory Limit: 65536K |
| Total Submissions: 19698 |
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Accepted: 8221 |
Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum. Input The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built. Output You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum. Sample Input 3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output 179 Source PKU Monthly,kicc |
题意：
有N个村庄，编号从1到N。现需要在这N个村庄之间修路，使得任何两个村庄之间都可以连通。称A、B两个村庄是连通的，当且仅当A与B有路直接连接，或者存在村庄C，使得A和C两村庄之间有路连接，且C和B之间有路连接。已知某些村庄之间已经有路直接连接了，试修建一些路使得所有村庄都是连通的、且修路总长度最短。
分析：
最小生成树问题，已经修好路的村庄之间将他们的长度置为0，然后再用Kruskal算法求解。
代码：
#include
#include
#include
#include
using namespace std; #define maxn 110 #define maxm 5050 int parent[maxn]; int g[maxn][maxn]; int m, ans; struct edge { int u, v, w; }EG[maxm]; bool cmp(edge a, edge b) { return a.w < b.w; } int Find(int x) { if(parent[x] == -1) return x; return Find(parent[x]); } void Kruskal() { memset(parent, -1, sizeof(parent)); ans = 0; sort(EG, EG+m, cmp); for(int i = 0; i < m; i++) { int t1 = Find(EG[i].u), t2 = Find(EG[i].v); if(t1 != t2) { ans += EG[i].w; parent[t1] = t2; } } } int main() { int n, q, a, b; while(~scanf("%d", &n)) { for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) scanf("%d", &g[i][j]); scanf("%d", &q); while(q--) { scanf("%d%d", &a, &b); g[a][b] = g[b][a] = 0; } m = 0; for(int i = 1; i <= n; i++) for(int j = i+1; j <= n; j++) { EG[m].u = i; EG[m].v = j; EG[m].w = g[i][j]; m++; } Kruskal(); printf("%d\n", ans); } return 0; }