📄 usaco 3_2_6 sweet butter 题解_leokan的blog.mht
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}
void heap_down(int i){
int a =3D 2*i+1;
int b =3D 2*i+2;
if(b < heapsize){
if(heap_val[b] < heap_val[a] && heap_val[b] < =
heap_val[i]){
heap_swap(i, b);
heap_down(b);
return;
}
}
if(a < heapsize && heap_val[a] < heap_val[i]){
heap_swap(i, a);
heap_down(a);
}
}
int main(){
FILE *filein =3D fopen("butter.in", "r");
fscanf(filein, "%d %d %d", &cows, &v, &e);
for(int i =3D 0; i < cows; ++i){
fscanf(filein, "%d", &cow_pos[i]);
--cow_pos[i];
}
for(int i =3D 0; i < v; ++i){
degree[i] =3D 0;
}
for(int i =3D 0; i < e; ++i){
int a,b,c;
fscanf(filein, "%d %d %d", &a, &b, &c);
--a;
--b;
con[a][degree[a]] =3D b;
cost[a][degree[a]] =3D c;
++degree[a];
con[b][degree[b]] =3D a;
cost[b][degree[b]] =3D c;
++degree[b];
}
fclose(filein);
for(int i =3D 0; i < cows; ++i){
heapsize =3D v;
for(int j =3D 0; j < v; ++j){
heap_id[j] =3D j;
heap_val[j] =3D BIG;
heap_lookup[j] =3D j;
}
heap_val[cow_pos[i]] =3D 0;
heap_up(cow_pos[i]);
bool fixed[MAXV];
memset(fixed, false, v);
for(int j =3D 0; j < v; ++j){
int p =3D heap_id[0];
dist[i][p] =3D heap_val[0];
fixed[p] =3D true;
heap_swap(0, heapsize-1);
--heapsize;
heap_down(0);
for(int k =3D 0; k < degree[p]; ++k){
int q =3D con[p][k];
if(!fixed[q]){
if(heap_val[heap_lookup[q]] > dist[i][p] + cost[p][k]){
heap_val[heap_lookup[q]] =3D dist[i][p] + cost[p][k];
heap_up(heap_lookup[q]);
}
}
}
}
}
int best =3D BIG;
for(int i =3D 0; i < v; ++i){
int total =3D 0;
for(int j =3D 0; j < cows; ++j){
total +=3D dist[j][i];
}
best <?=3D total;
}
FILE *fileout =3D fopen("butter.out", "w");
fprintf(fileout, "%d\n", best);
fclose(fileout);
return(0);
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