POJ 2421 Constructing Roads 最小生成树
题意:还是给你n个点,然后求最小生成树。特殊之处在于有一些点之间已经连上了边。思路:对于已经有边的点,特殊标记一下,加边的时候把这些边的权值赋值为0即可。这样就可以既保证这些边一定存在,又保证了所求的结果正确。
代码:
[cpp]
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <climits>
using namespace std;
#define CLR(arr,val) memset(arr,val,sizeof(arr))
const int N = 110;
int father[N];
struct edge{
int lp,rp,value;
}ee[N*N];
int map[N][N],flag[N][N],numedge,n;
bool cmp(edge a,edge b){
return a.value < b.value;
}
int find(int x){
if(x == father[x])
return father[x];
return find(father[x]);
}
bool Union_Set(int x,int y){
int fx = find(x);
int fy = find(y);
if(fx == fy){
return false;
}
else{
father[fx] = fy;
return true;
}
}
int kruskal(){
sort(ee,ee+numedge,cmp);
for(int i = 1; i <= n; ++i)
father[i] = i;
int sum = 0;
for(int i = 0; i < numedge; ++i){
int lx = ee[i].lp;
int rx = ee[i].rp;
if(Union_Set(lx,rx)){
sum += ee[i].value;
}
}
return sum;
}
int main(){
//freopen("1.txt","r",stdin);
while(scanf("%d",&n) != EOF){
CLR(flag,0);
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= n; ++j)
scanf("%d",&map[i][j]);
}
int m,x,y;
scanf("%d",&m);
while(m--){
scanf("%d%d",&x,&y);
flag[x][y] = 1;
}
numedge = 0;
for(int i = 1; i < n; ++i){
for(int j = i + 1; j <= n; ++j){
if(flag[i][j]){
ee[numedge].lp = i;
ee[numedge].rp = j;
ee[numedge].value = 0;
numedge++;
}
else{
ee[numedge].lp = i;
ee[numedge].rp = j;
ee[numedge].value = map[i][j];
numedge++;
}
}
}
int ans = kruskal();
printf("%d\n",ans);
}
return 0;
}
补充:软件开发 , C++ ,