題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=2433
題意:一個由n個點、m條邊構成的圖,問分別去掉每個邊後,剩下的圖任意兩點間最短路的和,如果不邊通輸出“INF"。
分析:先從每一個點求出它的單源最短路生成樹(road數組),同時可以求出從此點到所有點最短路的和(sum數組),然後枚舉每一條邊,如果點s的最短路生成樹中含有這條邊,就需要重新求s點的最短路生成樹,判斷時可以通過road數組O(1)時間判斷出來,結果就是sum( sum[1……n] );
可以用一個二維數組map[][]來判斷重邊,這樣在枚舉邊的時候如果是重邊,去掉這條邊肯定不影響結果,直接輸出sum數組的和或”INF"。
參考代碼:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define CLR(arr,v) memset(arr,v,sizeof(arr))
struct Node{
int num,next,id;
}ed[6005];
int h[105],pos;
int road[105][105],sum[105],dis[105],S[3005],T[3005];
bool vis[105],flag;
int map[105][105];
int m,n;
void init()
{
pos = 0;
flag = false;
CLR(h,-1);
CLR(road,-1);
CLR(sum,0);
CLR(map,0);
}
void add(int u,int v,int id)
{
ed[pos].num = v;
ed[pos].id = id;
ed[pos].next = h[u];
h[u] = pos++;
}
int Q[105],head,total;
void bfs(int s)
{
CLR(vis,false);
head = total = 0;
Q[total++] = s;
vis[s] = true;
dis[s] = 0;
int cnt = 0;
while(head != total)
{
int cur = Q[head++];
cnt++;
for(int i = h[cur];i != -1;i = ed[i].next)
{
int t = ed[i].num;
if(!vis[t])
{
vis[t] = true;
dis[t] = dis[cur] + 1;
road[s][t] = cur;
sum[s] += dis[t];
Q[total++] = t;
}
}
}
if(cnt != n) flag = true;
}
int bfs_del(int s,int id)
{
head = total = 0;
CLR(vis,false);
Q[total++] = s;
vis[s] = true;
dis[s] = 0;
int cnt = 0,sum1 = 0;
while(head != total)
{
int cur = Q[head++];
cnt++;
for(int i = h[cur];i != -1;i = ed[i].next)
{
int t = ed[i].num;
if((!vis[t]) && ed[i].id != id)
{
vis[t] = true;
dis[t] = dis[cur] + 1;
sum1 += dis[t];
Q[total++] = t;
}
}
}
if(cnt != n) return -1;
else return sum1;
}
void solve()
{
int total_sum = 0;
for(int i = 1;i <= n;++i)
{
bfs(i);
total_sum += sum[i];
}
int res,s,t;
for(int i = 0;i < m;++i)
{
res = 0;
s = S[i];
t = T[i];
if(flag)
{
printf("INF\n");
continue;
}
if(map[s][t] == 1)
{
for(int j = 1;j <= n;++j)
{
if(road[j][s] != t && road[j][t] != s)
res += sum[j];
else
{
int ans = bfs_del(j,i);
if(ans == -1)
{
res = -1;
break;
}
res += ans;
}
}
}
else res = total_sum;
if(res == -1) printf("INF\n");
else printf("%d\n",res);
}
}
int main()
{
while(scanf("%d%d",&n,&m) != EOF)
{
init();
int u,v;
for(int i = 0;i < m;++i)
{
scanf("%d%d",&S[i],&T[i]);
map[ S[i] ][ T[i] ] = ++map[ T[i] ][ S[i] ];
if(map[ S[i] ][ T[i] ] > 1) continue;
add(S[i],T[i],i);
add(T[i],S[i],i);
}
solve();
}
return 0;
}