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| Time Limit: 1000MS | Memory Limit: 131072K | |
| Total Submissions: 6646 | Accepted: 1788 |
Description
Lily runs a repairing company that services the Q blocks in the city. One day the company receives M repair tasks, the ith of which occurs in block pi, has a deadline ti on any repairman’s arrival, which is also its starting time, and takes a single repairman di time to finish. Repairmen work alone on all tasks and must finish one task before moving on to another. With a map of the city in hand, Lily want to know the minimum number of repairmen that have to be assign to this day’s tasks.
Input
The input contains multiple test cases. Each test case begins with a line containing Q and M (0 < Q ≤ 20, 0 < M ≤ 200). Then follow Q lines each with Q integers, which represent a Q × Q matrix Δ = {δij}, where δij means a bidirectional road connects the ith and the jth blocks and requires δij time to go from one end to another. If δij = −1, such a road does not exist. The matrix is symmetric and all its diagonal elements are zeroes. Right below the matrix are M lines describing the repairing tasks. The ith of these lines contains pi, ti and di. Two zeroes on a separate line come after the last test case.
Output
For each test case output one line containing the minimum number of repairmen that have to be assigned.
Sample Input
1 2 0 1 1 10 1 5 10 0 0
Sample Output
2
Source
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <vector> 6 #include <queue> 7 #include <cmath> 8 #include <set> 9 using namespace std; 10 11 #define N 25 12 #define M 205 13 14 int max(int x,int y){return x>y?x:y;} 15 int min(int x,int y){return x<y?x:y;} 16 int abs(int x,int y){return x<0?-x:x;} 17 18 struct node{ 19 int p, t, d; 20 }a[M]; 21 22 int map[N][N]; 23 int map2[N][N]; 24 vector<int>ve[M]; 25 int from[M]; 26 bool visited[M]; 27 int n; 28 29 int march(int u){ 30 int i, v; 31 for(i=0;i<ve[u].size();i++){ 32 v=ve[u][i]; 33 if(!visited[v]){ 34 visited[v]=true; 35 if(from[v]==-1||march(from[v])){ 36 from[v]=u; 37 return 1; 38 } 39 } 40 } 41 return 0; 42 } 43 44 45 main() 46 { 47 int i, j, k; 48 int q; 49 while(scanf("%d %d",&n,&q)==2){ 50 if(n==0&&q==0) break; 51 for(i=1;i<=n;i++){ 52 for(j=1;j<=n;j++){ 53 scanf("%d",&map[i][j]); 54 } 55 } 56 for(i=0;i<q;i++) scanf("%d %d %d",&a[i].p,&a[i].t,&a[i].d); 57 for(i=1;i<=n;i++){ 58 for(j=1;j<=n;j++){ 59 for(k=1;k<=n;k++){ 60 if(map[j][i]!=-1&&map[i][k]!=-1){ 61 if(map[j][k]==-1) map[j][k]=map[j][i]+map[i][k]; 62 else map[j][k]=min(map[j][k],map[j][i]+map[i][k]); 63 } 64 } 65 } 66 } 67 for(i=0;i<q;i++) ve[i].clear(); 68 for(i=0;i<q;i++){ 69 for(j=0;j<q;j++){ 70 if(map[a[i].p][a[j].p]!=-1&&a[i].t+a[i].d+map[a[i].p][a[j].p]<=a[j].t){ 71 ve[i].push_back(j); 72 // printf("1111\n"); 73 } 74 } 75 } 76 int num=0; 77 memset(from,-1,sizeof(from)); 78 for(i=0;i<q;i++){ 79 memset(visited,false,sizeof(visited)); 80 if(march(i)) num++; 81 } 82 // printf("%d\n",num); 83 printf("%d\n",q-num); 84 } 85 }
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原文地址：http://www.cnblogs.com/qq1012662902/p/4641855.html
