Journey
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://acm.uestc.edu.cn/#/problem/show/92Description
Bob has traveled to byteland, he find the N cities in byteland formed a tree structure, a tree structure is very special structure, there is exactly one path connecting each pair of nodes, and a tree with N nodes has N−1 edges.
As a traveler, Bob wants to journey between those N cities, and he know the time each road will cost. he advises the king of byteland building a new road to save time, and then, a new road was built. Now Bob has Q journey plan, give you the start city and destination city, please tell Bob how many time is saved by add a road if he always choose the shortest path. Note that if it's better not journey from the new roads, the answer is 0.
Input
First line of the input is a single integer T(1≤T≤20), indicating there are T test cases.
For each test case, the first will line contain two integers N(2≤N≤105) and Q(1≤Q≤105), indicating the number of cities in byteland and the journey plans. Then N line followed, each line will contain three integer x, y(1≤x,y≤N) and z(1≤z≤1000) indicating there is a road cost z time connect the xth city and the yth city, the first N−1 roads will form a tree structure, indicating the original roads, and the Nth line is the road built after Bob advised the king. Then Q line followed, each line will contain two integer x and y(1≤x,y≤N), indicating there is a journey plan from the xth city to yth city.
Output
For each case, you should first output Case #t: in a single line, where t indicating the case number between 1 and T, then Q lines followed, the ith line contains one integer indicating the time could saved in ith journey plan.
Sample Input
1 5 5 1 2 3 2 3 4 4 1 5 3 5 1 3 1 5 1 2 1 3 2 5 3 4 4 5
Sample Output
Case #1: 0 2 0 2 2
HINT
题意
给你一棵树，然后加了一条边，然后给Q次询问，问你这些点之间的最短距离缩短了多少
题解：
加了边之后，你走的方式就变成三种了，要么和原来一样，要么就是u-A-B-v,要么就是u-B-A-v这种
tarjan预处理一下距离跑一发就好了
@)1%KBO0HM418$J94$1R.jpg)
代码：
//qscqesze #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <bitset> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 200006 #define mod 1000000007 #define eps 1e-9 #define e exp(1.0) #define PI acos(-1) const double EP = 1E-10 ; int Num; //const int inf=0x7fffffff; const ll inf=999999999; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //*********************************************************** struct ndoe{ int v,w,next; }ed[maxn*2]; int dp[18][maxn*2],pos[maxn],dis[maxn],res[maxn],head[maxn],parent[maxn],vis[maxn]; int n,m,c,num,cnt,size; void addedge(int u,int v,int w) { ed[num].v=v; ed[num].w=w; ed[num].next=head[u]; head[u]=num++; } int Find(int i) { if(i!=parent[i]) parent[i]=Find(parent[i]); return parent[i]; } void Union(int i,int j) { int x,y; x=Find(i); y=Find(j); if(x!=y) parent[x]=y; } int A,B,C,q; void init() { int i,j,k; n=read();q=read(); m=n-1; memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); for(i=0;i<=n;i++) parent[i]=i; cnt=size=num=0; while(m--) { scanf("%d%d%d",&i,&j,&k); addedge(i,j,k); addedge(j,i,k); Union(i,j); } A=read(),B=read(),C=read(); } void dfs(int u,int dist) { int i,j; vis[u]=1; dis[u]=dist; pos[u]=cnt; res[size]=u; dp[0][cnt++]=size++; for(i=head[u];i!=-1;i=ed[i].next) { j=ed[i].v; if(!vis[j]) { dfs(j,dist+ed[i].w); dp[0][cnt++]=dp[0][pos[u]]; } } } void rmq() { int i,j,k; for(i=1;(1<<i)<=n;i++) for(j=n-1;j>=0;j--) { k=(1<<(i-1)); dp[i][j]=dp[i-1][j]; if(k+j<n) dp[i][j]=min(dp[i][j],dp[i-1][j+k]); } } int cal(int i,int j) { int k; if(i<j) { i^=j; j^=i; i^=j; } k=0; while((1<<k)<=(i-j+1)) k++; k--; k=min(dp[k][j],dp[k][i-(1<<k)+1]); return res[k]; } int Dis(int u,int v) { int k = cal(pos[u],pos[v]); return dis[u]+dis[v]-dis[k]*2; } int tot = 0; void solve() { int i,j,k; for(i=1;i<=n;i++) if(!vis[i]) dfs(i,0); n=n*2-1; rmq(); printf("Case #%d: ",tot); for(int i=1;i<=q;i++) { int u=read(),v=read(); int P = Dis(u,v); int PP1 = Dis(u,A)+C+Dis(B,v); int PP2 = Dis(u,B)+C+Dis(A,v); PP1 = min(PP1,PP2); printf("%d ",max(P-PP1,0)); } } int main() { int t=read(); while(t--) { tot++; init(); solve(); } return 0; }