最长k可重区间集问题【网络流24题】
作者：互联网
思路
由要求线段的长度，很容易想到应该把问题转化成求费用流。
通过限制好相邻点之间的流量，就能保证每个区间内保证不会有使用次数超过x次的点。
然后再把区间作为主要要求的目标，把一个区间看作一个有点权的点连在图中。
因为区间只能使用一次，且为了计算长度，我们让这个区间的费用为 -len。
这样跑MCMF所得出的mincost就是跑满图时得到的最大区间长度。
但是由于本题数据给出的区间范围相当大
正常的建图只能过掉前六个点
考虑到 n 很小，说明区间大时不需要用的点很多
所以这些点之间连边与否是完全不影响整张图的流量的
这时考虑离散化，就能在找残量网络和更新最大费用时减少毫不相干的时空支出
CODE
1 #include <bits/stdc++.h>
2 #define dbg(x) cout << #x << "=" << x << endl
3 #define eps 1e-8
4 #define pi acos(-1.0)
5
6 using namespace std;
7 typedef long long LL;
8 const int maxn = 1e5 +7;
9 const int inf = 0x3f3f3f3f;
10
11 int n, m, s, t;
12 int head[maxn],pre[maxn],inq[maxn],dis[maxn];
13 int a[maxn];
14 int cnt = 1;
15 int path[2][maxn];
16 int mincost = 0, maxflow = 0;
17 struct edge {
18 int u,to,nxt,w,c;
19 }e[maxn << 1];
20 int tot[5];
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 inline void BuildGraph(int u, int v, int w, int cost)
30 {
31 e[++cnt] = (edge){u, v, head[u], w, cost}, head[u] = cnt;
32 e[++cnt] = (edge){v, u, head[v], 0, -cost}, head[v] = cnt;///反向边
33 }
34
35 queue<int> q;
36
37 inline void init() {
38 memset(head, 0, sizeof(head));
39 memset(pre, 0, sizeof(pre));
40 memset(inq, 0, sizeof(inq));
41 memset(dis, 0, sizeof(dis));
42 memset(e, 0, sizeof(e));
43 while(!q.empty()) {
44 q.pop();
45 }
46 mincost = maxflow = 0;
47 cnt = 1;
48 }
49
50 bool SPFA(int x)
51 {
52 memset(inq, 0, sizeof(inq));
53 for(int i = s; i <= t; i++) {
54 dis[i] = inf;
55 }
56 q.push(x);
57 dis[x] = 0;
58 inq[x] = 1;
59 while(!q.empty()) {
60 int u = q.front();
61 q.pop();
62 inq[u] = 0;
63 for(int i = head[u]; i; i = e[i].nxt) {
64 int v = e[i].to, w = e[i].c;
65 if(e[i].w > 0) {
66 if(dis[u] + w < dis[v]) {
67 dis[v] = dis[u] + w;
68 pre[v] = i;
69 if(!inq[v]) {
70 q.push(v);
71 inq[v] = 1;
72 }
73 }
74 }
75 }
76 }
77 if(dis[t] == inf)
78 return 0;
79 return 1;
80 }
81
82 void MCMF()
83 {
84 while(SPFA(s)) {
85 int temp = inf;
86 for(int i = pre[t]; i; i = pre[e[i].u]) {
87 temp = min(temp, e[i].w);
88 }
89 for(int i = pre[t]; i; i = pre[e[i].u]) {
90 e[i].w -= temp;
91 e[i^1].w += temp;
92 mincost += e[i].c * temp;
93 //printf("e[%d].c:%d\n",i, e[i].c);
94 //printf("ans:%d\n",ans);
95 }
96 //maxflow += temp;
97 }
98 }
99
100 int k;
101 int l[maxn], r[maxn];
102 int ls[maxn], lss[maxn];
103 int totls, totlss;
104 int len[maxn];
105
106 void Unique() {
107 for ( int i = 1; i <= n; ++i ) {
108 for ( int j = 1; j <= totlss; ++j ) {
109 if(l[i] == lss[j]) {
110 l[i] = j;
111 }
112 if(r[i] == lss[j]) {
113 r[i] = j;
114 }
115 }
116 }
117 }
118
119 int main()
120 {
121 //freopen("data.txt", "r", stdin);
122 read(n); read(k);
123 init();
124 for ( int i = 1; i <= n; ++i ) {
125 read(l[i]); read(r[i]);
126 ls[++totls] = l[i];
127 ls[++totls] = r[i];
128 len[i] = r[i] - l[i];
129 }
130 sort(ls + 1, ls + 1 + totls);
131 for ( int i = 1; i <= totls; ++i ) {
132 if(ls[i] == ls[i-1])
133 continue;
134 lss[++totlss] = ls[i];
135 }
136 Unique();
137 s = 0;
138 for ( int i = 1; i <= n; ++i ) {
139 t = max(t, r[i]);
140 BuildGraph(l[i], r[i], 1, -len[i]);
141 }
142 for ( int i = 0; i <= t; ++i ) {
143 BuildGraph(i, i + 1, k, 0);
144 }
145 ++t;
146 MCMF();
147 cout << -mincost << endl;
148 return 0;
149 }
View Code
#include <bits/stdc++.h> #define dbg(x) cout << #x << "=" << x << endl #define eps 1e-8 #define pi acos(-1.0)
using namespace std; typedef long long LL; const int maxn = 1e5 +7; const int inf = 0x3f3f3f3f;
int n, m, s, t; int head[maxn],pre[maxn],inq[maxn],dis[maxn]; int a[maxn]; int cnt = 1; int path[2][maxn]; int mincost = 0, maxflow = 0; struct edge { int u,to,nxt,w,c; }e[maxn << 1]; int tot[5];
template<class T>inline void read(T &res) { char c;T flag=1; while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; }
inline void BuildGraph(int u, int v, int w, int cost) { e[++cnt] = (edge){u, v, head[u], w, cost}, head[u] = cnt; e[++cnt] = (edge){v, u, head[v], 0, -cost}, head[v] = cnt;///反向边 }
queue<int> q;
inline void init() { memset(head, 0, sizeof(head)); memset(pre, 0, sizeof(pre)); memset(inq, 0, sizeof(inq)); memset(dis, 0, sizeof(dis)); memset(e, 0, sizeof(e)); while(!q.empty()) { q.pop(); } mincost = maxflow = 0; cnt = 1; }
bool SPFA(int x) { memset(inq, 0, sizeof(inq)); for(int i = s; i <= t; i++) { dis[i] = inf; } q.push(x); dis[x] = 0; inq[x] = 1; while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = 0; for(int i = head[u]; i; i = e[i].nxt) { int v = e[i].to, w = e[i].c; if(e[i].w > 0) { if(dis[u] + w < dis[v]) { dis[v] = dis[u] + w; pre[v] = i; if(!inq[v]) { q.push(v); inq[v] = 1; } } } } } if(dis[t] == inf) return 0; return 1; }
void MCMF() { while(SPFA(s)) { int temp = inf; for(int i = pre[t]; i; i = pre[e[i].u]) { temp = min(temp, e[i].w); } for(int i = pre[t]; i; i = pre[e[i].u]) { e[i].w -= temp; e[i^1].w += temp; mincost += e[i].c * temp; //printf("e[%d].c:%d\n",i, e[i].c); //printf("ans:%d\n",ans); } //maxflow += temp; } }
int k; int l[maxn], r[maxn]; int ls[maxn], lss[maxn]; int totls, totlss; int len[maxn];
void Unique() { for ( int i = 1; i <= n; ++i ) { for ( int j = 1; j <= totlss; ++j ) { if(l[i] == lss[j]) { l[i] = j; } if(r[i] == lss[j]) { r[i] = j; } } } }
int main() { //freopen("data.txt", "r", stdin); read(n); read(k); init(); for ( int i = 1; i <= n; ++i ) { read(l[i]); read(r[i]); ls[++totls] = l[i]; ls[++totls] = r[i]; len[i] = r[i] - l[i]; } sort(ls + 1, ls + 1 + totls); for ( int i = 1; i <= totls; ++i ) { if(ls[i] == ls[i-1]) continue; lss[++totlss] = ls[i]; } Unique(); s = 0; for ( int i = 1; i <= n; ++i ) { t = max(t, r[i]); BuildGraph(l[i], r[i], 1, -len[i]); } for ( int i = 0; i <= t; ++i ) { BuildGraph(i, i + 1, k, 0); } ++t; MCMF(); cout << -mincost << endl; return 0; }
标签：24,pre,可重,temp,int,memset,maxn,sizeof,最长 来源： https://www.cnblogs.com/orangeko/p/12820125.html