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
由於老尼克，在又重新演算法。
定一根，找到有向最小生成。
Prim 和 Kruskal 是派不上用的，但多少有。
首先，基於婪的思路，先根排除在外，
其他找到各自的最小重入，明地方式，
符合需求，每一非根一定要有入，才能有解。
然而在才，如果有怎？
考成一，指向的重稍微修改一番，
仔思考一下，v 在上(w 也是)，原本定 wv 最小，
在找到一次小的 uv (u 在外面)，的重修正
uv-wv，假使我到一，最小的成本最增加
uv-wv，而情是 wv 移除，入 uv，拆。
#include <stdio.h>
#include <vector>
#include <string.h>
#include <math.h>
using namespace std;
struct Node {
int x, y;
double v;// x->y, v
int next;
} edge[100005];
int e, g[105], vg[105];// direct graph record
void addEdge(int x, int y, double v) {
if(x == y) return;
edge[e].x = x, edge[e].y = y, edge[e].v = v;
edge[e].next = g[x], g[x] = e++;
edge[e].x = y, edge[e].y = x, edge[e].v = -127;//this weight not important.
edge[e].next = vg[y], vg[y] = e++;
}
int visited[105];
void dfs(int x) {
visited[x] = 1;
for(int i = g[x]; i != -1; i = edge[i].next) {
if(!visited[edge[i].y])
dfs(edge[i].y);
}
}
double mst(int root, int n) {//node [0, n-1]
#define oo 1000000000
int i, j, k;
int u;
&nbs; double v;
memset(visited, 0, sizeof(visited));
dfs(root);
for(i = 0; i < n; i++)
if(visited[i] == 0)
return -1;
// solvable
int prev[105], dele[105];
double ret = 0;
double vprev[105];
memset(dele, 0, sizeof(dele));
while(true) {
for(i = 0; i < n; i++)
prev[i] = i, vprev[i] = oo;
for(i = 0; i < n; i++) {// O(E)
if(dele[i])
continue;
for(j = g[i]; j != -1; j = edge[j].next) {
u = edge[j].y, v = edge[j].v;
if(dele[u])
continue;
if(vprev[u] > v)
vprev[u] = v, prev[u] = i;
}
}
for(i = 0; i < n; i++) {
if(dele[i] || i == root)
continue;
j = i; // find cycle
memset(visited, 0, sizeof(visited));
visited[root] = 1;
do visited[j] = 1, j = prev[j];
while(!visited[j]);
if(j == root) continue;
i = j;
ret += vprev[i];
for(j = prev[i]; j != i; j = prev[j]) {
ret += vprev[j];
dele[j] = 1;
}
// modify edge weight.
//printf("modify edge weight\n");
for(j = vg[i]; j != -1; j = edge[j].next) {
u = edge[j].y;// u->i
if(dele[u])
continue;
edge[j^1].v -= vprev[i];
}
double in[105], out[105];
for(k = 0; k < n; k++)
in[k] = oo, out[k] = oo;
for(j = prev[i]; j != i; j = prev[j]) {
for(k = vg[j]; k != -1; k = edge[k].next) {
u = edge[k].y, v = edge[k^1].v;// u->cycle
if(dele[u])
continue;
v -= vprev[j];
in[u] = min(in[u], v);
}
for(k = g[j]; k != -1; k = edge[k].next) {
u = edge[k].y, v = edge[k].v;// cycle->u
if(dele[u])
continue;
out[u] = min(out[u], v);
}
}
for(k = 0; k < n; k++) {
if(k == i)
continue;
if(in[k] != oo)
addEdge(k, i, in[k]);
if(out[k] != oo)
addEdge(i, k, out[k]);
}
break;// repeat for a new graph.
}
if(i == n) {// there not exist cycle.
for(i = 0; i < n; i++) {
if(dele[i] || i == root)
continue;
//printf("%lf\n", vprev[i]);
ret += vprev[i];
}
break;
}
}
return ret;
}
int main() {
int n, m;
int i, j, k, u, v;
int x[105], y[105];
while(scanf("%d %d", &n, &m) == 2) {
e = 0;
memset(g, -1, sizeof(g));
memset(vg, -1, sizeof(vg));
for(i = 0; i < n; i++)
scanf("%d %d", &x[i], &y[i]);
for(i = 0; i < m; i++) {
scanf("%d %d", &u, &v);
u--, v--;
addEdge(u, v, hypot(x[u]-x[v], y[u]-y[v]));
// printf("%d %d - %lf\n", u, v, hypot(x[u]-x[v], y[u]-y[v]));
}
double ret = mst(0, n);
if(ret < 0)
puts("poor snoopy");
else
printf("%.2f\n", ret);
}
return 0;
}
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