POJ 3216 Repairing Company【Floyd + 最小路径覆盖】

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <vector>
 5 using namespace std;
 6 
 7 const int maxn = 25;
 8 const int INF = 1000000000;
 9 int mat[maxn][maxn];
10 struct Task {
11     int block, start, time;
12 }task[205];
13 
14 void Floyd(int n) {
15     for(int k = 1; k <= n; k++) {
16         for(int i = 1; i <= n; i++) {
17             for(int j = 1; j <= n; j++) {
18                 if(mat[i][k] + mat[k][j] < mat[i][j]) {
19                     mat[i][j] = mat[i][k] + mat[k][j];
20                 }
21             }
22         }
23     }
24 }
25 
26 vector<int> G[205];
27 int vis[205];
28 int Link[205];
29 bool Find(int u) {
30     for(int i = 0; i < G[u].size(); i++) {
31         int v = G[u][i];
32         if(!vis[v]) {
33             vis[v] = 1;
34             if(Link[v] == -1 || Find(Link[v]) ) {
35                 Link[v] = u;
36                 return true;
37             }
38         }
39     }
40     return false;
41 }
42 
43 int solve(int m) {
44     int cnt = 0;
45     memset(Link, -1, sizeof(Link));
46     for(int i = 0; i < m; i++) {
47         if(G[i].size()) {
48             memset(vis, 0, sizeof(vis));
49             if(Find(i)) cnt++;
50         }
51     }
52     return cnt;
53 }
54 
55 bool check(int i, int j) {
56     int need_time = mat[task[i].block][task[j].block];
57     if(need_time < INF) {
58         if(task[i].start + task[i].time + need_time <= task[j].start) {
59             return true;
60         }
61     }
62     return false;
63 }
64 
65 int main() {
66     int n, m;
67     while(scanf("%d %d",&n, &m) && n + m) {
68         for(int i = 1; i <= n; i++) {
69             for(int j = 1; j <= n; j++) {
70                 scanf("%d",&mat[i][j]);
71                 if(mat[i][j] == -1) mat[i][j] = INF;
72             }
73         }
74         Floyd(n);
75         for(int i = 0; i < m; i++) {
76             G[i].clear();
77             scanf("%d %d %d",&task[i].block, &task[i].start, &task[i].time);
78         }
79         for(int i = 0; i < m; i++) {
80             for(int j = 0; j < m; j++) {
81                 if(i == j) continue;
82                 if(check(i, j) ) G[i].push_back(j);
83             }
84         }
85         printf("%d\n", m - solve(m));
86     }
87     return 0;
88 }