最长k可重区间集问题网络流24题
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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 ",i, e[i].c); 94 //printf("ans:%d ",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 }
#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
",i, e[i].c);
//printf("ans:%d
",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;
}
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