HDU-2121-Ice_cream’s world II

描述

题解

代码

#include <iostream>
#include <cstring>

using namespace std;

/*
*  最小树形图
*  int型
*  复杂度O(NM)
*  点从0开始
*/
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 11010;

struct Edge
{
int u, v, cost;
};

Edge edge[MAXM];

int pos;
int pre[MAXN], id[MAXN], visit[MAXN], in[MAXN];

int zhuliu(int root, int n, int m)
{
int res = 0, v;
while (1)
{
memset(in, 0x3f, sizeof(in));

for (int i = 0; i < m; i++)
{
if (edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v])
{
pre[edge[i].v] = edge[i].u;
in[edge[i].v] = edge[i].cost;
if (edge[i].u == root)
{
pos = i;
}
}
}
for (int i = 0; i < n; i++)
{
if (i != root && in[i] == INF)
{
return -1;  //  不存在最小树形图
}
}
int tn = 0;
memset(id, -1, sizeof(id));
memset(visit, -1, sizeof(visit));
in[root] = 0;
for (int i = 0; i < n; i++)
{
res += in[i];
v = i;
while (visit[v] != i && id[v] == -1 && v != root)
{
visit[v] = i;
v = pre[v];
}
if (v != root && id[v] == -1)
{
for (int u = pre[v]; u != v ; u = pre[u])
{
id[u] = tn;
}
id[v] = tn++;
}
}
if (tn == 0)
{
break;  //  没有有向环
}
for (int i = 0; i < n; i++)
{
if (id[i] == -1)
{
id[i] = tn++;
}
}
for (int i = 0; i < m; i++)
{
v = edge[i].v;
edge[i].u = id[edge[i].u];
edge[i].v = id[edge[i].v];
if (edge[i].u != edge[i].v)
{
edge[i].cost -= in[v];
}
}
n = tn;
root = id[root];
}
return res;
}

int main()
{
int n, m;
while (~scanf("%d%d", &n, &m))
{
int sum = 0;
for (int i = 0; i < m; i++)
{
scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].cost);
edge[i].u++;
edge[i].v++;
sum += edge[i].cost;
}
sum++;
//  增加超级节点0,节点0到其余各个节点的边权相同（此题中 边权要大于原图的总边权值）
for (int i = m; i < m + n; i++)
{
edge[i].u = 0;
edge[i].v = i - m + 1;
edge[i].cost = sum;
}
int ans = zhuliu(0, n + 1, m + n);
//  n+1为总结点数,m+n为总边数
//  ans代表以超级节点0为根的最小树形图的总权值,
//  将ans减去sum,如果差值小于sum,说明节点0的出度只有1,说明原图是连通图
//  如果差值>=sum,那么说明节点0的出度不止为1,说明原图不是连通图
if (ans == -1 || ans - sum >= sum)
{
puts("impossible");
}
else
{
printf("%d %d\n",ans - sum, pos - m);
}
puts("");
}

return 0;
}

参考