Command Network | Time Limit: 1000MS | | Memory Limit: 131072K | | Total Submissions: 7267 | | Accepted: 2160 |
Description After a long lasting war on words, a war on arms finally breaks out between littleken’s and KnuthOcean’s kingdoms. A sudden and violent assault by KnuthOcean’s force has rendered a total failure of littleken’s command network. A provisional network must be built immediately. littleken orders snoopy to take charge of the project. With the situation studied to every detail, snoopy believes that the most urgent point is to enable littenken’s commands to reach every disconnected node in the destroyed network and decides on a plan to build a unidirectional communication network. The nodes are distributed on a plane. If littleken’s commands are to be able to be delivered directly from a node A to another node B, a wire will have to be built along the straight line segment connecting the two nodes. Since it’s in wartime, not between all pairs of nodes can wires be built. snoopy wants the plan to require the shortest total length of wires so that the construction can be done very soon. Input The input contains several test cases. Each test case starts with a line containing two integer N (N ≤ 100), the number of nodes in the destroyed network, and M (M ≤ 104), the number of pairs of nodes between which a wire can be built. The next N lines each contain an ordered pair xi and yi, giving the Cartesian coordinates of the nodes. Then follow M lines each containing two integers i and j between 1 and N (inclusive) meaning a wire can be built between node i and node j for unidirectional command delivery from the former to the latter. littleken’s headquarter is always located at node 1. Process to end of file. Output For each test case, output exactly one line containing the shortest total length of wires to two digits past the decimal point. In the cases that such a network does not exist, just output ‘poor snoopy’. Sample Input 4 6
0 6
4 6
0 0
7 20
1 2
1 3
2 3
3 4
3 1
3 2
4 3
0 0
1 0
0 1
1 2
1 3
4 1
2 3 Sample Output 31.19
poor snoopy Source |
裸的最小樹形圖，我就是為了檢驗?zāi)０?...結(jié)果還錯了.....%.2lf在POJ上用G++會莫名的掛掉，所以用C++就行了。
代碼：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define MAXN 110
#define INF 1e30
using namespace std;
inline double sqr(double x)
{
return x * x;
}
struct point
{
double _x, _y;
double dist_with(const point &rh) const
{
return sqrt(sqr(_x - rh._x) + sqr(_y - rh._y));
}
} po[MAXN];
double g[MAXN][MAXN];
int pre[MAXN];
bool is_out[MAXN];
int vis[MAXN], vcnt;
double solve(int n, int root)
{
memset(is_out, false, n);
double ans = 0;
while (1)
{
int i, j, k;
for (i = 0; i < n; ++i)
if (i != root && !is_out[i])
{
pre[i] = i;
for (j = 0; j < n; ++j)
if (!is_out[j] && g[pre[i]][i] > g[j][i])
pre[i] = j;
if (pre[i] == i) throw false;
}
for (i = 0; i < n; ++i)
if (i != root && !is_out[i])
{
j = i;
vis[i] = ++vcnt;
while (vis[pre[j]] != vcnt && pre[j] != root)
{
j = pre[j];
vis[j] = vcnt;
}
if (pre[j] == i)
break;
}
if (i == n)
{
for (j = 0; j < n; ++j)
if (j != root && !is_out[j])
ans += g[pre[j]][j];
break;
}
j = i;
do
{
is_out[j] = true;
ans += g[pre[j]][j];
j = pre[j];
} while (j != i);
for (int j = 0; j < n; ++j)
if (vis[j] == vcnt)
for (int k = 0; k < n; ++k)
if (!is_out[k])
{
if (g[i][k] > g[j][k])
g[i][k] = g[j][k];
if (g[k][i] > g[k][j] - g[pre[j]][j] && g[k][j] != INF)
g[k][i] = g[k][j] - g[pre[j]][j];
}
is_out[i] = false;
}
return ans;
}
int main()
{
int n, m;
while (scanf("%d%d", &n, &m) != EOF)
{
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
g[i][j] = INF;
for (int i = 0; i < n; ++i)
scanf("%lf%lf", &po[i]._x, &po[i]._y);
for (int i = 0; i < m; ++i)
{
int a, b;
scanf("%d%d", &a, &b);
if (a != b)
g[a - 1][b - 1] = po[a - 1].dist_with(po[b - 1]);
}
try
{
printf("%.2lf\n", solve(n, 0));
}
catch (bool)
{
printf("poor snoopy\n");
}
}
return 0;
}
posted on 2011-10-15 22:18
LLawliet 閱讀(163)
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