Bonsai
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 802 Accepted Submission(s): 406
Problem Description
After being assaulted in the parking lot by Mr. Miyagi following the "All Valley Karate Tournament", John Kreese has come to you for assistance. Help John in his quest for justice by chopping off all the leaves from Mr. Miyagi‘s bonsai
tree!
You are given an undirected tree (i.e., a connected graph with no cycles), where each edge (i.e., branch) has a nonnegative weight (i.e., thickness). One vertex of the tree has been designated the root of the tree.The remaining vertices of the tree each have
unique paths to the root; non-root vertices which are not the successors of any other vertex on a path to the root are known as leaves.Determine the minimum weight set of edges that must be removed so that none of the leaves in the original tree are connected
by some path to the root.
Input
The input file will contain multiple test cases. Each test case will begin with a line containing a pair of integers n (where 1 <= n <= 1000) and r (where r ∈ ｛1,……, n｝) indicating the number of vertices in the tree and the index
of the root vertex, respectively. The next n-1 lines each contain three integers ui vi wi (where ui, vi ∈ {1,……, n} and 0 <= wi <= 1000) indicating that vertex ui is connected to
vertex vi by an undirected edge with weight wi. The input file will not contain duplicate edges. The end-of-file is denoted by a single line containing "0 0".
Output
For each input test case, print a single integer indicating the minimum total weight of edges that must be deleted in order to ensure that there exists no path from one of the original leaves to the root.
Sample Input
15 15 1 2 1 2 3 2 2 5 3 5 6 7 4 6 5 6 7 4 5 15 6 15 10 11 10 13 5 13 14 4 12 13 3 9 10 8 8 9 2 9 11 3 0 0
Sample Output
16
Source
2009 Stanford Local ACM Programming Contest
题意：给一棵n个节点的树，根节点为 r ，每条边都有一个花费。现在要求断开所有的叶子节点到根节点 r 的路径，问删除的边的花费最小是多少。
解题：最小割，源点VS＝r ，新增一个汇点VT＝n+1，叶子节点与汇点相连，边容为INF。其他的边按原输入建图就可以了。
/*
最大流：SAP算法，与ISAP的差别就是不用预处理
*/
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define captype int
const int MAXN = 1010; //点的总数
const int MAXM = 400010; //边的总数
const int INF = 1<<30;
struct EDG{
int to,next;
captype cap,flow;
} edg[MAXM];
int eid,head[MAXN];
int gap[MAXN]; //每种距离(或可认为是高度)点的个数
int dis[MAXN]; //每个点到终点eNode 的最短距离
int cur[MAXN]; //cur[u] 表示从u点出发可流经 cur[u] 号边
int pre[MAXN];
void init(){
eid=0;
memset(head,-1,sizeof(head));
}
//有向边 三个参数，无向边4个参数
void addEdg(int u,int v,captype c,captype rc=0){
edg[eid].to=v; edg[eid].next=head[u];
edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++;
edg[eid].to=u; edg[eid].next=head[v];
edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;
}
captype maxFlow_sap(int sNode,int eNode, int n){//n是包括源点和汇点的总点个数，这个一定要注意
memset(gap,0,sizeof(gap));
memset(dis,0,sizeof(dis));
memcpy(cur,head,sizeof(head));
pre[sNode] = -1;
gap[0]=n;
captype ans=0; //最大流
int u=sNode;
while(dis[sNode]<n){ //判断从sNode点有没有流向下一个相邻的点
if(u==eNode){ //找到一条可增流的路
captype Min=INF ;
int inser;
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]) //从这条可增流的路找到最多可增的流量Min
if(Min>edg[i].cap-edg[i].flow){
Min=edg[i].cap-edg[i].flow;
inser=i;
}
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){
edg[i].flow+=Min;
edg[i^1].flow-=Min; //可回流的边的流量
}
ans+=Min;
u=edg[inser^1].to;
continue;
}
bool flag = false; //判断能否从u点出发可往相邻点流
int v;
for(int i=cur[u]; i!=-1; i=edg[i].next){
v=edg[i].to;
if(edg[i].cap-edg[i].flow>0 && dis[u]==dis[v]+1){
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if(flag){
u=v;
continue;
}
//如果上面没有找到一个可流的相邻点，则改变出发点u的距离（也可认为是高度）为相邻可流点的最小距离+1
int Mind= n;
for(int i=head[u]; i!=-1; i=edg[i].next)
if(edg[i].cap-edg[i].flow>0 && Mind>dis[edg[i].to]){
Mind=dis[edg[i].to];
cur[u]=i;
}
gap[dis[u]]--;
if(gap[dis[u]]==0) return ans; //当dis[u]这种距离的点没有了，也就不可能从源点出发找到一条增广流路径
//因为汇点到当前点的距离只有一种，那么从源点到汇点必然经过当前点，然而当前点又没能找到可流向的点，那么必然断流
dis[u]=Mind+1;//如果找到一个可流的相邻点，则距离为相邻点距离+1，如果找不到，则为n+1
gap[dis[u]]++;
if(u!=sNode) u=edg[pre[u]^1].to; //退一条边
}
return ans;
}
int main()
{
int n,VS ,VT ,u,v,w , d[MAXN];
while(scanf("%d%d",&n,&VS)>0&&n+VS!=0){
init();
memset(d,0,sizeof(d));
VT=n+1;
for(int i=1; i<n; i++){
scanf("%d%d%d",&u,&v,&w);
d[u]++; d[v]++;
addEdg(u,v,w);
addEdg(v,u,w);
}
for(int i=1; i<=n; i++)
if(i!=VS&&d[i]==1)
addEdg(i,VT,INF);
printf("%d\n",maxFlow_sap(VS,VT,VT));
}
}
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