poj 3680 Intervals 费用流
给一些线段， 每个线段有一个值， 并且覆盖一些点， 求每个点被覆盖次数不超过k时， 可以取得的最大值。
首先将点离散化， 然后连边， i向i+1连一条容量为k， 费用为0的边。 对于每条线段， 起点向终点连一条容量为1， 费用为-val的边， 然后跑费用流就好。
1 #include <iostream> 2 #include <vector> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <cmath> 7 #include <map> 8 #include <set> 9 #include <string> 10 #include <queue> 11 #include <stack> 12 #include <bitset> 13 using namespace std; 14 #define pb(x) push_back(x) 15 #define ll long long 16 #define mk(x, y) make_pair(x, y) 17 #define lson l, m, rt<<1 18 #define mem(a) memset(a, 0, sizeof(a)) 19 #define rson m+1, r, rt<<1|1 20 #define mem1(a) memset(a, -1, sizeof(a)) 21 #define mem2(a) memset(a, 0x3f, sizeof(a)) 22 #define rep(i, n, a) for(int i = a; i<n; i++) 23 #define fi first 24 #define se second 25 typedef pair<int, int> pll; 26 const double PI = acos(-1.0); 27 const double eps = 1e-8; 28 const int mod = 1e9+7; 29 const int inf = 1061109567; 30 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; 31 const int maxn = 2e5+5; 32 int num, head[maxn*2], s, t, n, k, nn, dis[maxn], flow, cost, cnt, cap[maxn], q[maxn], cur[maxn], vis[maxn]; 33 struct node 34 { 35 int to, nextt, c, w; 36 node(){} 37 node(int to, int nextt, int c, int w):to(to), nextt(nextt), c(c), w(w) {} 38 }e[maxn*2]; 39 int spfa() { 40 int st, ed; 41 st = ed = 0; 42 mem2(dis); 43 ++cnt; 44 dis[s] = 0; 45 cap[s] = inf; 46 cur[s] = -1; 47 q[ed++] = s; 48 while(st<ed) { 49 int u = q[st++]; 50 vis[u] = cnt-1; 51 for(int i = head[u]; ~i; i = e[i].nextt) { 52 int v = e[i].to, c = e[i].c, w = e[i].w; 53 if(c && dis[v]>dis[u]+w) { 54 dis[v] = dis[u]+w; 55 cap[v] = min(c, cap[u]); 56 cur[v] = i; 57 if(vis[v] != cnt) { 58 vis[v] = cnt; 59 q[ed++] = v; 60 } 61 } 62 } 63 } 64 if(dis[t] == inf) 65 return 0; 66 cost += dis[t]*cap[t]; 67 flow += cap[t]; 68 for(int i = cur[t]; ~i; i = cur[e[i^1].to]) { 69 e[i].c -= cap[t]; 70 e[i^1].c += cap[t]; 71 } 72 return 1; 73 } 74 int mcmf() { 75 flow = cost = 0; 76 while(spfa()) 77 ; 78 return cost; 79 } 80 void add(int u, int v, int c, int val) { 81 e[num] = node(v, head[u], c, val); head[u] = num++; 82 e[num] = node(u, head[v], 0, -val); head[v] = num++; 83 } 84 void init() { 85 mem1(head); 86 num = cnt = 0; 87 mem(vis); 88 } 89 pair <int, pll > a[205]; 90 int b[450]; 91 int main() 92 { 93 ios::sync_with_stdio(0); 94 int T, m, x, y, z; 95 cin>>T; 96 while(T--) { 97 cin>>n>>m; 98 init(); 99 int cnt = 0; 100 for(int i = 0; i<n; i++) { 101 scanf("%d%d%d", &a[i].fi, &a[i].se.fi, &a[i].se.se); 102 b[cnt++] = a[i].fi, b[cnt++] = a[i].se.fi; 103 } 104 sort(b, b+cnt); 105 cnt = unique(b, b+cnt)-b; 106 for(int i = 1; i<=cnt; i++) { 107 add(i, i+1, m, 0); 108 } 109 for(int i = 0; i<n; i++) { 110 int u = lower_bound(b, b+cnt, a[i].fi)-b+1; 111 int v = lower_bound(b, b+cnt, a[i].se.fi)-b+1; 112 add(u, v, 1, -a[i].se.se); 113 } 114 s = 0, t = cnt+1; 115 add(s, 1, m, 0); 116 add(cnt, t, m, 0); 117 int ans = -mcmf(); 118 cout<<ans<<endl; 119 } 120 return 0; 121 }
