题意:要在n个城市之间建造公路,使城市之间能互相联通,告诉每个城市之间建公路的费用,和已经建好的公路,求最小费用。
解法:最小生成树。先把已经建好的边加进去再跑kruskal或者prim什么的。
代码:
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long using namespace std; struct node
{
int u, v, val;
bool operator < (const node &tmp) const
{
return val < tmp.val;
}
}edge[10005];
int father[105];
int Find(int a)
{
if(a != father[a]) father[a] = Find(father[a]);
return father[a];
}
bool Union(int a, int b)
{
int c = Find(a), d = Find(b);
if(c == d) return false;
father[c] = d;
return true;
}
int main()
{
int n;
while(~scanf("%d", &n))
{
int cnt = 0;
for(int i = 1; i <= n; i++)
father[i] = i;
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
int x;
scanf("%d", &x);
if(i <= j) continue;
edge[cnt].u = i;
edge[cnt].v = j;
edge[cnt++].val = x;
}
}
int m;
scanf("%d", &m);
for(int i = 0; i < m; i++)
{
int a, b;
scanf("%d%d", &a, &b);
Union(a, b);
}
sort(edge, edge + cnt);
int ans = 0;
for(int i = 0; i < cnt; i++)
{
if(Union(edge[i].u, edge[i].v)) ans += edge[i].val;
}
printf("%d\n", ans);
}
return 0;
}