PAT-Top1001. Battle Over Cities - Hard Version (35)

  1 //#include "stdafx.h"
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <vector>
  5 
  6 using namespace std;
  7 
  8 struct edge { // edge struct
  9     int u, v, cost;
 10 };
 11 vector<edge> edges; // the number of edges is greater than 500 far and away
 12 
 13 int cmp(edge a, edge b) { // sort rule
 14     return a.cost < b.cost;
 15 }
 16 
 17 int parent[510]; // union-find set
 18 
 19 void initParent(int n) { // initialize union-find set
 20     int i;
 21     for(i = 1; i <= n; i++) {
 22         parent[i] = -1; // a minus means it is a root node and its absolute value represents the number of the set
 23     }
 24 }
 25 
 26 int findRoot(int x) { // find the root of the set
 27     int s = x;
 28     while(parent[s] > 0) {
 29         s = parent[s];
 30     }
 31 
 32     int temp;
 33     while(s != x) { // compress paths for fast lookup
 34         temp = parent[x];
 35         parent[x] = s;
 36         x = temp;
 37     }
 38 
 39     return s;
 40 }
 41 
 42 void unionSet(int r1, int r2) { // union sets. More concretely, merge a small number of set into a large collection
 43     int sum = parent[r1] + parent[r2];
 44     if(parent[r1] > parent[r2]) { 
 45         parent[r1] = r2;
 46         parent[r2] = sum;
 47     } else {
 48         parent[r2] = r1;
 49         parent[r1] = sum;
 50     }
 51 }
 52 
 53 int maxw = 1; // max cost
 54 bool infw; // infinite cost
 55 
 56 int kruskal(int n, int m, int out) { // Kruskal algorithm to get minimum spanning tree
 57     initParent(n);
 58 
 59     int u, v, r1, r2, num = 0, i, w = 0;
 60     for (i = 0; i < m; i++) {
 61         u = edges[i].u;
 62         v = edges[i].v;
 63 
 64         if (u == out || v == out) {
 65             continue;
 66         }
 67 
 68         r1 = findRoot(u);
 69         r2 = findRoot(v);
 70 
 71         if (r1 != r2) {
 72             unionSet(r1, r2);
 73             num++;
 74 
 75             if (edges[i].cost > 0) { // only consider the cost which is not zero
 76                 w += edges[i].cost;
 77             }
 78 
 79             if (num == n - 2) {
 80                 break;
 81             }
 82         }
 83     }
 84 
 85     //printf("num %d\n", num);
 86     if (num < n - 2) { // spanning tree is not connected
 87         w = -1; // distinguish the situation of the occurrence of infinite cost
 88 
 89         if (!infw) { // when infinite cost first comes out
 90             infw = true;
 91         }
 92     }
 93 
 94     return w;
 95 }
 96 
 97 int main() {
 98     int n, m;
 99     scanf("%d%d", &n, &m);
100 
101     int i, status;
102     edge e;
103     for (i = 0; i < m; i++) {
104         scanf("%d%d%d%d", &e.u, &e.v, &e.cost, &status);
105         if (status == 1) {
106             e.cost = 0;
107         }
108 
109         edges.push_back(e);
110     }
111 
112     if (m > 0) {
113         sort(edges.begin(), edges.end(), cmp);
114     }
115 
116     int curw, res[510], index = 0;
117     for (i = 1; i <= n; i++) { // traverse all vertices to obtain the target vertex
118         curw = kruskal(n, m, i);
119         if (!infw) { // when infinite cost doesn‘t come out
120             if (curw < maxw) {
121                 continue;
122             }
123 
124             if (curw > maxw) {
125                 index = 0;
126                 maxw = curw;
127             } 
128             res[index++] = i;
129         } else { // otherwise
130             if (curw < 0) {
131                 if (maxw > 0) {
132                     maxw = -1;
133                     index = 0;
134                 } 
135 
136                 res[index++] = i;
137             }
138         }
139     }
140 
141     if (index > 0) {
142         for (i = 0; i < index; i++) {
143             if (i > 0) {
144                 printf(" ");
145             } 
146             printf("%d", res[i]);
147         }
148     } else {
149         printf("0");
150     }
151     printf("\n");
152 
153     system("pause");
154     return 0;
155 }