刷题总结——Genghis Khan the Conqueror (hdu4126)
题目：
Our story is about Jebei Noyan(哲别), who was one of the most famous generals in Genghis Khan’s cavalry. Once his led the advance troop to invade a country named Pushtuar. The knights rolled up all the cities in Pushtuar rapidly. As Jebei Noyan’s advance troop did not have enough soldiers, the conquest was temporary and vulnerable and he was waiting for the Genghis Khan’s reinforce. At the meantime, Jebei Noyan needed to set up many guarders on the road of the country in order to guarantee that his troop in each city can send and receive messages safely and promptly through those roads.
There were N cities in Pushtuar and there were bidirectional roads connecting cities. If Jebei set up guarders on a road, it was totally safe to deliver messages between the two cities connected by the road. However setting up guarders on different road took different cost based on the distance, road condition and the residual armed power nearby. Jebei had known the cost of setting up guarders on each road. He wanted to guarantee that each two cities can safely deliver messages either directly or indirectly and the total cost was minimal.
Things will always get a little bit harder. As a sophisticated general, Jebei predicted that there would be one uprising happening in the country sooner or later which might increase the cost (setting up guarders) on exactly ONE road. Nevertheless he did not know which road would be affected, but only got the information of some suspicious road cost changes. We assumed that the probability of each suspicious case was the same. Since that after the uprising happened, the plan of guarder setting should be rearranged to achieve the minimal cost, Jebei Noyan wanted to know the new expected minimal total cost immediately based on current information.
Input
There are no more than 20 test cases in the input.
For each test case, the first line contains two integers N and M (1<=N<=3000, 0<=M<=N×N), demonstrating the number of cities and roads in Pushtuar. Cities are numbered from 0 to N-1. In the each of the following M lines, there are three integers x i, y i and c i(c i<=10 7), showing that there is a bidirectional road between x i and y i, while the cost of setting up guarders on this road is c i. We guarantee that the graph is connected. The total cost of the graph is less or equal to 10 9.
The next line contains an integer Q (1<=Q<=10000) representing the number of suspicious road cost changes. In the following Q lines, each line contains three integers X i, Y i and C i showing that the cost of road (X i, Y i) may change to C i(C i<=10 7). We guarantee that the road always exists and C i is larger than the original cost (we guarantee that there is at most one road connecting two cities directly). Please note that the probability of each suspicious road cost change is the same.
Output
For each test case, output a real number demonstrating the expected minimal total cost. The result should be rounded to 4 digits after decimal point.
Sample Input
3 3 0 1 3 0 2 2 1 2 5 3 0 2 3 1 2 6 0 1 6 0 0
Sample Output
6.0000
Hint
The initial minimal cost is 5 by connecting city 0 to 1 and city 0 to 2. In the first suspicious case, the minimal total cost is increased to 6; the second case remains 5; the third case is increased to 7. As the result, the expected cost is (5+6+7)/3 = 6.
题解:
很好的一道树形dp题··
每个询问x,y其实求的就是相邻的两个子树x,y的最短距离··我们用best[x][y]表示
由于q很大··上述值肯定是通过预处理求出···首先求出最开始的最小生成树,接着我们要先求得f[i][j],表示以i为根节点,通过非生成树边到达j所在子树的最短距离···对此我们一一枚举0——n-1作为根节点然后树形dp即可求得···
求完f[i][j]的话best[x][y]就很简单了··我们只需枚举y所在子树的所有节点u··求出f[u][x]的最小值即可··最后再与新的增大的边c比较一下取最小值就可以了
代码：
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<ctime> #include<cctype> #include<algorithm> using namespace std; const int N=3005; const int M=9e6+5; const int inf=0x3f3f3f3f; inline int R() { char c;int f=0; for(c=getchar();c<'0'||c>'9';c=getchar()); for(;c<='9'&&c>='0';c=getchar()) f=(f<<3)+(f<<1)+c-'0'; return f; } struct node { int a,b,val; }ed[M]; int n,m,q,fst[N],nxt[N*2],go[N*2],val[N*2],tot,father[N],map[N][N],f[N][N],best[N][N]; double sum=0,ans=0; bool jud[N][N]; inline int get(int a) { if(father[a]==a) return a; else return father[a]=get(father[a]); } inline bool cmp(node a,node b) { return a.val<b.val; } inline void comb(int a,int b,int c) { nxt[++tot]=fst[a],fst[a]=tot,go[tot]=b,val[tot]=c; nxt[++tot]=fst[b],fst[b]=tot,go[tot]=a,val[tot]=c; } inline void pre() { tot=0;ans=sum=0; for(int i=0;i<n;i++) father[i]=i; memset(fst,0,sizeof(fst));memset(map,inf,sizeof(map)); memset(f,inf,sizeof(f));memset(jud,false,sizeof(jud)); memset(best,inf,sizeof(best)); } inline int dfs1(int u,int fa,int rt) { for(int e=fst[u];e;e=nxt[e]) { int v=go[e];if(v==fa) continue; f[rt][u]=min(f[rt][u],dfs1(v,u,rt)); } if(fa!=rt) f[rt][u]=min(f[rt][u],map[rt][u]); return f[rt][u]; } inline int dfs2(int u,int fa,int rt) { int ans=f[u][rt]; for(int e=fst[u];e;e=nxt[e]) { int v=go[e];if(v==fa) continue; ans=min(ans,dfs2(v,u,rt)); } return ans; } inline void dp() { for(int i=0;i<n;i++) dfs1(i,-1,i); for(int i=0;i<n;i++) for(int e=fst[i];e;e=nxt[e]) { int v=go[e];best[i][v]=best[v][i]=dfs2(v,i,i); } } int main() { // freopen("a.in","r",stdin); while(~scanf("%d%d",&n,&m)&&(n+m)) { int a,b,c;pre(); for(int i=1;i<=m;i++) { a=R(),b=R(),c=R();map[a][b]=map[b][a]=c; ed[i].a=a,ed[i].b=b,ed[i].val=c; } sort(ed+1,ed+m+1,cmp);int temp=0; for(int i=1;i<=m;i++) { int fa=get(ed[i].a),fb=get(ed[i].b); if(fa!=fb) { sum+=ed[i].val; father[fa]=fb;temp++; comb(ed[i].a,ed[i].b,ed[i].val); jud[ed[i].a][ed[i].b]=jud[ed[i].b][ed[i].a]=true; } if(temp==n-1) break; } dp(); q=R(); for(int t=1;t<=q;t++) { a=R(),b=R(),c=R(); if(!jud[a][b]) ans+=sum; else { int temp=min(c,best[a][b]); ans+=(sum-map[a][b]+temp); } } ans=(double)ans/q; printf("%0.4f\n",ans); } return 0; }
