Ice_cream’s world II
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2119 Accepted Submission(s): 502
Problem Description
After
awarded lands to ACMers, the queen want to choose a city be her
capital. This is an important event in ice_cream world, and it also a
very difficult problem, because the world have N cities and M roads,
every road was directed. Wiskey is a chief engineer in ice_cream world.
The queen asked Wiskey must find a suitable location to establish the
capital, beautify the roads which let capital can visit each city and
the project’s cost as less as better. If Wiskey can’t fulfill the
queen’s require, he will be punishing.
Input
Every
case have two integers N and M (N<=1000, M<=10000), the cities
numbered 0…N-1, following M lines, each line contain three integers S, T
and C, meaning from S to T have a road will cost C.
Output
If
no location satisfy the queen’s require, you must be output
“impossible”, otherwise, print the minimum cost in this project and
suitable city’s number. May be exist many suitable cities, choose the
minimum number city. After every case print one blank.
Sample Input
3 1
0 1 1
4 4
0 1 10
0 2 10
1 3 20
2 3 30
Sample Output
impossible
40 0
Author
Wiskey
Source
Recommend
威士忌
给定有向图，求最小树形图和根，用LRJ的模板WA了N次。下面AC代码照敲自：http://www.cnblogs.com/nanke/archive/2012/04/11/2441725.html
自己据理解加了注释：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<cmath>
#include<fstream>
#include<iomanip>
#include<climits>
#include<cfloat>
using namespace std;
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define ULL unsigned long long
#define LL long long
#define MAX(x,y) ((x) > (y) ? (x) : (y))
#define MIN(x,y) ((x) < (y) ? (x) : (y))
#define MAXN 1111
#define MAXM 11111
#define INF MAX_INT
int pre[MAXN],id[MAXN],visit[MAXN];
int inv[MAXN];
struct edge{
int u,v,d;
}e[MAXM];
int roote;
int mdst(int s, int n, int m){
int u,v,i,d;
int ret=0;
while(true){ //找出每个点的最小入弧
for(i=0; i<n; i++)
inv[i]=INF;
for(i=0; i<m; i++){
u=e[i].u;
v=e[i].v;
d=e[i].d;
if(d<inv[v] && u!=v){
inv[v]=d; pre[v]=u;
if(u==s) roote=i; //要求的根
}
} //当前点集合中有点不可达,则图不连通
for(i=0; i<n; i++) if(inv[i]==INF && i!=s)
return -1;
inv[s]=0;
int cnt=0;
memset(id, -1, sizeof(id));
memset(visit, -1, sizeof(visit));
for(i=0; i<n; i++){ //缩圈
ret+=inv[i];
v=i; //如果v还没标记且非虚根
while(visit[v]!=i && id[v]==-1 && v!=s){
visit[v]=i; v=pre[v];
} //这里表明上步visit[v]=i，即有圈
if(v!=s && id[v]==-1){
for(u=pre[v]; u!=v; u=pre[u])
id[u]=cnt;
id[v]=cnt++;
}
}
if(!cnt) break; //如果没圈
for(i=0; i<n; i++) if(id[i]==-1)
id[i]=cnt++;
for(i=0; i<m; i++){ //更新缩圈后各边
u=e[i].u; v=e[i].v; d=e[i].d;
e[i]=(edge){id[u], id[v], (u==v ? d : d-inv[v])};
}
n=cnt;
s=id[s];
}
return ret;
}
int main(){
//freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
int n,m;
while(scanf(" %d %d",&n,&m)==2){
int i,u_,v_,w_,s=1;
const int m1=m;
for(i=0; i<m; i++){
scanf(" %d %d %d",&u_,&v_,&w_);
e[i]=(edge){u_,v_,w_};
s+=w_;
}
for(i=0; i<n; i++)
e[m++]=(edge){n,i,s};
int ans=mdst(n, 1+n, m);
if(ans==-1 || ans-s>=s) printf("impossible
");
else printf("%d %d
",ans-s, roote-m1);
}
return 0;
}
为什么最后每条非自环都要减去最小入弧权值inv[v]？ ==因为缩圈时ret已经加上了inv[v]，
对所有e(u, v), e.d = e.d - inv[v]可保证下一次迭代为点v选出正确的最小入弧(inv[v]).