uva oj 567 - Risk(Floyd算法)

时间:2023-03-08 16:45:00
 /*
一张有20个顶点的图上。
依次输入每个点与哪些点直接相连。
并且多次询问两点间,最短需要经过几条路才能从一点到达另一点。 bfs 水过
*/
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<queue>
using namespace std; struct node{
int x, step;
node(){ }
node(int x, int step){
this->x=x;
this->step=step;
}
}; vector<int>v[];
queue<node>q;
int vis[]; int b, e; void bfs(){
while(!q.empty()) q.pop();
node cur;
q.push(node(b, ));
while(!q.empty()){
cur=q.front();
q.pop();
if(cur.x==e){
printf("%2d to %2d: %d\n", b, e, cur.step);
return;
}
int len=v[cur.x].size();
for(int i=; i<len; ++i){
if(v[cur.x][i]==e){
printf("%2d to %2d: %d\n", b, e, cur.step+);
return;
}
if(!vis[v[cur.x][i]]){
vis[v[cur.x][i]]=;
q.push(node(v[cur.x][i], cur.step+));
}
}
}
} int main(){
int n, u;
int cnt=;
while(scanf("%d", &n)!=EOF){
while(n--){
scanf("%d", &u);
v[].push_back(u);
v[u].push_back();
}
for(int i=; i<=; ++i){
scanf("%d", &n);
while(n--){
scanf("%d", &u);
v[i].push_back(u);
v[u].push_back(i);
}
}
scanf("%d", &n);
printf("Test Set #%d\n", ++cnt);
while(n--){
scanf("%d%d", &b, &e) ;
bfs();
memset(vis, , sizeof(vis));
}
printf("\n");
for(int i=; i<=; ++i)
v[i].clear(); }
return ;
}
 /*
Floyd 才是正解!
*/
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std; int map[][]; void Folyd(){
for(int k=; k<=; ++k)
for(int i=; i<=; ++i)
for(int j=; j<=; ++j)
if(map[i][j] > map[i][k] + map[k][j])
map[i][j] = map[i][k] + map[k][j];
} int main(){
int n, u, b, e;
int cnt=;
while(scanf("%d", &n)!=EOF){
memset(map, 0x3f, sizeof(map));
while(n--){
scanf("%d", &u);
map[][u]=map[u][]=;
}
for(int i=; i<=; ++i){
scanf("%d", &n);
while(n--){
scanf("%d", &u);
map[u][i]=map[i][u]=;
}
}
scanf("%d", &n);
printf("Test Set #%d\n", ++cnt);
Folyd();
while(n--){
scanf("%d%d", &b, &e) ;
printf("%2d to %2d: %d\n", b, e, map[b][e]);
}
printf("\n");
}
}