HDU1102并查集和最小生成树

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18668 Accepted Submission(s): 7114



Problem 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

题目大意：

要造路，这个路刚开始有连在一起的，也有分开的（间接连在一起的也算）。然后每条路之间造路都要一定的费用，这个费用用矩阵表示。

遇到的问题和思路：

prim算法中，如果两条路本来就连在一起的，那么used假设还是没有用过，但是cost[i][j]都是0，就好了

给出AC代码：

prim算法

#include
#include
#include

using namespace std;

const int inf = 0x3f3f3f3f;
const int MAX_N = 100 + 10;
const int MAX = inf * (inf + 1) / 2;
int n, q, a, b;
int cost[MAX_N][MAX_N];
int mincost[MAX_N];
bool used[MAX_N];

void work(){
for(int i = 1; i <= n; i++){
mincost[i] = inf;
}
mincost[1] = 0;
int res = 0;
while(true){
int v = -1;
for(int i = 1; i <= n; i++){
if(!used[i] && (v == -1 || mincost[i] < mincost[v]))v = i;
}
//printf("v = %d\n", v);
if(v == -1)break;
res += mincost[v];
used[v] = 1;
//printf("res = %d\n", res);
for(int i = 1; i <= n; i++){
mincost[i] = min(mincost[i], cost[v][i]);
}
}
printf("%d\n", res);
}

int main(){
while(scanf("%d%", &n)!=EOF){
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
scanf("%d", &cost[i][j]);
}
}
scanf("%d", &q);
memset(used, 0, sizeof(used));
for(int i = 1; i <= q; i++){
scanf("%d%d", &a, &b);
cost[a][b] = 0;
cost[b][a] = 0;
}
work();
}
return 0;
}