Intervals

Time Limit: 5000ms

Memory Limit: 65536KB

This problem will be judged on 

PKU. Original ID: 

64-bit integer IO format: %lld      Java class name: Main

 

You are given N weighted open intervals. The ith interval covers (ai, bi) and weighs wi. Your task is to pick some of the intervals to maximize the total weights under the limit that no point in the real axis is covered more than k times.

 

Input

The first line of input is the number of test case.The first line of each test case contains two integers, N and K (1 ≤ K ≤ N ≤ 200).The next N line each contain three integers ai, bi, wi(1 ≤ ai < bi ≤ 100,000, 1 ≤ wi ≤ 100,000) describing the intervals. There is a blank line before each test case.

 

Output

For each test case output the maximum total weights in a separate line.

 

Sample Input

43 11 2 22 3 43 4 83 11 3 22 3 43 4 83 11 100000 1000001 2 3100 200 3003 21 100000 1000001 150 301100 200 300

Sample Output

1412100000100301

Source

 

解题：离散化+费用流

 

1 #include 
        
           2 #include 
         
            3 #include 
          
             4 #include 
           
              5 #include 
            
               6 #include 
             
               7 #include 
              
                8 #include 
               
                 9 #include 
                
                  10 #include 
                 
                   11 #include 
                  
                    12 #include 
                   
                     13 #define LL long long 14 #define INF 0x3f3f3f3f 15 using namespace std; 16 const int maxn = 1000; 17 struct arc{ 18 int v,w,f,next; 19 arc(int x = 0,int y = 0,int z = 0,int nxt = 0){ 20 v = x; 21 w = y; 22 f = z; 23 next = nxt; 24 } 25 }; 26 arc e[10000]; 27 int head[maxn],d[maxn],p[maxn],tot; 28 int n,k,S,T,lisan[maxn],cnt,x[maxn],y[maxn],sc[maxn]; 29 bool in[maxn]; 30 queue
                    
                     q; 31 void add(int u,int v,int w,int f){ 32 e[tot] = arc(v,w,f,head[u]); 33 head[u] = tot++; 34 e[tot] = arc(u,-w,0,head[v]); 35 head[v] = tot++; 36 } 37 bool spfa(){ 38 for(int i = 0; i <= T; i++){ 39 d[i] = INF; 40 in[i] = false; 41 p[i] = -1; 42 } 43 while(!q.empty()) q.pop(); 44 d[S] = 0; 45 in[S] = true; 46 q.push(S); 47 while(!q.empty()){ 48 int u = q.front(); 49 q.pop(); 50 in[u] = false; 51 for(int i = head[u]; ~i; i = e[i].next){ 52 if(e[i].f > 0 && d[e[i].v] > d[u] + e[i].w){ 53 d[e[i].v] = d[u] + e[i].w; 54 //cout<<"end:"<
                     
                      <
                      
                        -1; 64 } 65 int solve(){ 66 int tmp = 0,minV; 67 while(spfa()){ 68 minV = INF; 69 for(int i = p[T]; ~i; i = p[e[i^1].v]) 70 minV = min(minV,e[i].f); 71 for(int i = p[T]; ~i; i = p[e[i^1].v]){ 72 tmp += minV*e[i].w; 73 e[i].f -= minV; 74 e[i^1].f += minV; 75 } 76 //cout<<"tmp:"<
                       
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