Ice_cream’s world II
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 938 Accepted Submission(s): 192
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
Author
Wiskey
Source
Recommend
威士忌
思路：不定根的最小树形图，做法就是虚拟一个根，让所有点到这个根连线边，并且边权值很大，这样就能保证最后只有一个点连向这个虚拟根，这样最后结果减去这个很大值就行了。
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 1010
#define MAXE 10100
using namespace std;
typedef long long typec;
const typec INF = 0x7fffffffffffffffLL;
struct Arcs
{
int _a, _b;
} arcs[MAXE + MAXN];
struct AdjNode
{
int _v;
Arcs *_rf;
typec _c;
AdjNode *_next;
} *adj[MAXN], adjmem[MAXE + MAXN];
int adjcnt;
inline void init(int n)
{
memset(adj, 0, n * sizeof(AdjNode *));
adjcnt = 0;
}
inline void add_edge(int a, int b, typec c)
{
arcs[adjcnt]._a = a;
arcs[adjcnt]._b = b;
AdjNode *ptr = adjmem + adjcnt++;
ptr -> _c = c;
ptr -> _v = a;
ptr -> _rf = arcs + adjcnt - 1;
ptr -> _next = adj[b];
adj[b] = ptr;
}
int pre[MAXN], real_pre[MAXN];
bool is_out[MAXN];
int vis[MAXN], vcnt;
typec solve(int n, int root)
{
static typec ch[MAXN];
memset(is_out, false, n);
typec ans = 0;
while (1)
{
int i, j, k;
for (i = 0; i < n; ++i)
if (i != root && !is_out[i])
{
pre[i] = i;
ch[i] = INF;
AdjNode *chp;
for (AdjNode *ptr = adj[i]; ptr; ptr = ptr -> _next)
{
j = ptr -> _v;
if (!is_out[j])
{
if (ch[i] > ptr -> _c)
{
pre[i] = j;
ch[i] = ptr -> _c;
chp = ptr;
}
else if (ch[i] == ptr -> _c && ptr -> _rf -> _a == root && ptr -> _rf -> _b < chp -> _rf -> _b)
{
pre[i] = j;
chp = ptr;
}
}
}
if (pre[i] == i) throw false;
real_pre[chp -> _rf -> _b] = chp -> _rf -> _a;
}
memset(vis, 0, n * sizeof(int));
vcnt = 0;
for (i = 0; i < n; ++i)
if (i != root && !is_out[i])
{
j = i;
vis[i] = ++ vcnt;
while (vis[pre[j]] == 0 && pre[j] != root)
{
j = pre[j];
vis[j] = vcnt;
}
if (vis[pre[j]] == vcnt)
{
i = pre[j];
break;
}
}
if (i == n)
{
for (j = 0; j < n; ++j)
if (j != root && !is_out[j])
ans += ch[j];
break;
}
j = i;
++vcnt;
int ti = i;
do
{
ti = min(ti, j);
vis[j] = vcnt;
is_out[j] = true;
ans += ch[j];
j = pre[j];
}
while (j != i);
i = ti;
for (j = 0; j < n; ++j)
if (vis[j] == vcnt)
for (AdjNode **ptr = adj + j; *ptr;)
{
k = (*ptr) -> _v;
if (!is_out[k])
{
AdjNode *p2 = *ptr;
p2 -> _c -= ch[j];
if (i != j)
{
*ptr = p2 -> _next;
p2 -> _next = adj[i];
adj[i] = p2;
}
else
ptr = &((*ptr) -> _next);
}
else
ptr = &((*ptr) -> _next);
}
for (k = 0; k < n; ++k)
if (!is_out[k])
for (AdjNode *ptr = adj[k]; ptr; ptr = ptr -> _next)
{
j = ptr -> _v;
if (vis[j] == vcnt)
ptr -> _v = i;
}
is_out[i] = false;
}
return ans;
}
int main()
{
int n, m;
while (scanf("%d%d", &n, &m) != EOF)
{
init(n + 1);
long long s = 1;
for (int i = 0; i < m; ++i)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if (a != b)
{
add_edge(a, b, c);
s += c;
}
}
for (int i = 0; i < n; ++i)
add_edge(n, i, s);
long long ans = solve(n + 1, n);
int r, p;
for (r = 0; real_pre[r] != n; ++r);
for (p = r + 1; p < n; ++p)
if (real_pre[p] == n)
break;
if (p == n)
printf("%I64d %d\n\n", ans - s, r);
else
printf("impossible\n\n");
}
return 0;
}