输入输出样例
输入 #1
4
输出 #1
View Code
11 1 8 2 7 9 3 6 10 4 5 11
思路
既然是网络流24题自然用网络流的思想来建图
考虑一个问题,
如果把每个柱子考虑成一个特定路径,
把柱子上的球看作是路径经过的点号,
为了要在固定的路线上经过更多的点,
则可以很清晰的知道这是一个最小路径覆盖问题。
看出了方向之后自然是思考建图:
大家都清楚网络流擅长解决此类有限制条件的线性规划问题。
自然要先分析如何把限制条件转化成图上点与点之间的关系上去。
首先,先枚举球的个数。竟然没有T
把点逐渐加到路径上,
因为每个点既要和S连也要和T连,
还要和其他点权和为完全平方数的点相连,
自然考虑拆点。
通过画图容易发现:
1、S -> i入, i出 -> T,容量为1
2、i入 -> j出
当不断加点把图跑满时,再加点不会更新最大流
这时就应该引入新的路径,也就是增加柱子个数
直到柱子个数 = n
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 9 const int inf = 0x3f3f3f3f; 10 11 template<class T>inline void read(T &res) 12 { 13 char c;T flag=1; 14 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; 15 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; 16 } 17 18 namespace _buff { 19 const size_t BUFF = 1 << 19; 20 char ibuf[BUFF], *ib = ibuf, *ie = ibuf; 21 char getc() { 22 if (ib == ie) { 23 ib = ibuf; 24 ie = ibuf + fread(ibuf, 1, BUFF, stdin); 25 } 26 return ib == ie ? -1 : *ib++; 27 } 28 } 29 30 int qread() { 31 using namespace _buff; 32 int ret = 0; 33 bool pos = true; 34 char c = getc(); 35 for (; (c < '0' || c > '9') && c != '-'; c = getc()) { 36 assert(~c); 37 } 38 if (c == '-') { 39 pos = false; 40 c = getc(); 41 } 42 for (; c >= '0' && c <= '9'; c = getc()) { 43 ret = (ret << 3) + (ret << 1) + (c ^ 48); 44 } 45 return pos ? ret : -ret; 46 } 47 48 const int maxn = 200007; 49 50 int n, m; 51 int s, t; 52 53 struct edge{ 54 int from,to; 55 LL cap,flow; 56 }; 57 58 int Pre[maxn << 1], Nxt[maxn << 1]; 59 bool vis[maxn << 1]; 60 61 struct DINIC { 62 int head[maxn << 1], nxt[maxn << 1], edge[maxn << 1], cnt; 63 int cap[maxn << 1], depth[maxn << 1]; 64 65 void init() { 66 cnt = 1; 67 memset(head, 0, sizeof(head)); 68 } 69 70 void BuildGraph(int u, int v, int w) { 71 ++cnt; 72 edge[cnt] = v; 73 nxt[cnt] = head[u]; 74 cap[cnt] = w; 75 head[u] = cnt; 76 77 ++cnt; 78 edge[cnt] = u; 79 nxt[cnt] = head[v]; 80 cap[cnt] = 0; 81 head[v] = cnt; 82 } 83 84 queue<int> q; 85 86 bool bfs() { 87 memset(depth, 0, sizeof(depth)); 88 depth[s] = 1; 89 q.push(s); 90 while(!q.empty()) { 91 int u = q.front(); 92 q.pop(); 93 for ( int i = head[u]; i; i = nxt[i] ) { 94 int v = edge[i]; 95 if(depth[v]) { 96 continue; 97 } 98 if(cap[i]) { 99 depth[v] = depth[u] + 1; 100 q.push(v); 101 } 102 } 103 } 104 return depth[t]; 105 } 106 107 int dfs(int u, int dist) { 108 if(u == t) { 109 return dist; 110 } 111 int flow = 0; 112 for ( int i = head[u]; i && dist; i = nxt[i] ) { 113 if(cap[i] == 0) 114 continue; 115 int v = edge[i]; 116 if(depth[v] != depth[u] + 1) { 117 continue; 118 } 119 int res = dfs(v, min(cap[i], dist)); 120 cap[i] -= res; 121 cap[i ^ 1] += res; 122 //printf("cap[%d]:%d ",t, cap[t]); 123 dist -= res; 124 flow += res; 125 Nxt[u / 2] = v / 2; 126 } 127 return flow; 128 } 129 130 int maxflow() { 131 int ans = 0; 132 while(bfs()) { 133 ans += dfs(s, inf); 134 } 135 return ans; 136 } 137 } dinic; 138 139 int main() 140 { 141 //freopen("data.txt", "r", stdin); 142 read(n); 143 dinic.init(); 144 s = 0, t = 1e5 + 7; 145 int balls = 0, coll = 0; 146 while(coll <= n) { 147 ++balls; 148 dinic.BuildGraph(s, balls * 2, 1); 149 dinic.BuildGraph(balls * 2 + 1, t, 1); 150 for ( int i = sqrt(balls) + 1; i * i < (balls * 2); ++i ) { 151 dinic.BuildGraph((i * i - balls) * 2, balls * 2 + 1, 1); 152 } 153 int maxflow = dinic.maxflow(); 154 if(maxflow == 0) { 155 ++coll; 156 Pre[coll] = balls; 157 } 158 } 159 printf("%d ",balls - 1); 160 for ( int i = 1; i <= n; ++i ) { 161 if(!vis[Pre[i]]) { 162 for ( int u = Pre[i]; u && u != (t / 2); u = Nxt[u] ) { 163 vis[u] = true; 164 printf("%d ",u); 165 } 166 puts(""); 167 } 168 } 169 return 0; 170 }
#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 inf = 0x3f3f3f3f;
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;
}
namespace _buff {
const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
if (ib == ie) {
ib = ibuf;
ie = ibuf + fread(ibuf, 1, BUFF, stdin);
}
return ib == ie ? -1 : *ib++;
}
}
int qread() {
using namespace _buff;
int ret = 0;
bool pos = true;
char c = getc();
for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
assert(~c);
}
if (c == '-') {
pos = false;
c = getc();
}
for (; c >= '0' && c <= '9'; c = getc()) {
ret = (ret << 3) + (ret << 1) + (c ^ 48);
}
return pos ? ret : -ret;
}
const int maxn = 200007;
int n, m;
int s, t;
struct edge{
int from,to;
LL cap,flow;
};
int Pre[maxn << 1], Nxt[maxn << 1];
bool vis[maxn << 1];
struct DINIC {
int head[maxn << 1], nxt[maxn << 1], edge[maxn << 1], cnt;
int cap[maxn << 1], depth[maxn << 1];
void init() {
cnt = 1;
memset(head, 0, sizeof(head));
}
void BuildGraph(int u, int v, int w) {
++cnt;
edge[cnt] = v;
nxt[cnt] = head[u];
cap[cnt] = w;
head[u] = cnt;
++cnt;
edge[cnt] = u;
nxt[cnt] = head[v];
cap[cnt] = 0;
head[v] = cnt;
}
queue<int> q;
bool bfs() {
memset(depth, 0, sizeof(depth));
depth[s] = 1;
q.push(s);
while(!q.empty()) {
int u = q.front();
q.pop();
for ( int i = head[u]; i; i = nxt[i] ) {
int v = edge[i];
if(depth[v]) {
continue;
}
if(cap[i]) {
depth[v] = depth[u] + 1;
q.push(v);
}
}
}
return depth[t];
}
int dfs(int u, int dist) {
if(u == t) {
return dist;
}
int flow = 0;
for ( int i = head[u]; i && dist; i = nxt[i] ) {
if(cap[i] == 0)
continue;
int v = edge[i];
if(depth[v] != depth[u] + 1) {
continue;
}
int res = dfs(v, min(cap[i], dist));
cap[i] -= res;
cap[i ^ 1] += res;
//printf("cap[%d]:%d
",t, cap[t]);
dist -= res;
flow += res;
Nxt[u / 2] = v / 2;
}
return flow;
}
int maxflow() {
int ans = 0;
while(bfs()) {
ans += dfs(s, inf);
}
return ans;
}
} dinic;
int main()
{
//freopen("data.txt", "r", stdin);
read(n);
dinic.init();
s = 0, t = 1e5 + 7;
int balls = 0, coll = 0;
while(coll <= n) {
++balls;
dinic.BuildGraph(s, balls * 2, 1);
dinic.BuildGraph(balls * 2 + 1, t, 1);
for ( int i = sqrt(balls) + 1; i * i < (balls * 2); ++i ) {
dinic.BuildGraph((i * i - balls) * 2, balls * 2 + 1, 1);
}
int maxflow = dinic.maxflow();
if(maxflow == 0) {
++coll;
Pre[coll] = balls;
}
}
printf("%d
",balls - 1);
for ( int i = 1; i <= n; ++i ) {
if(!vis[Pre[i]]) {
for ( int u = Pre[i]; u && u != (t / 2); u = Nxt[u] ) {
vis[u] = true;
printf("%d ",u);
}
puts("");
}
}
return 0;
}