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Codeforces Round #576 (Div. 1) 简要题解 (CDEF)
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1198 C Matching vs Independent Set
大意: 给定$3n$个点的无向图, 求构造$n$条边的匹配, 或$n$个点的独立集.
假设已经构造出$x$条边的匹配, 那么剩余$3n-2x$个点, 若$x<n$, 则$3n-2x\ge n$可以构造出独立集.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head const int N = 1e6+;
int u[N], v[N], a[N], vis[N]; void work() {
int n, m;
scanf("%d%d", &n, &m);
int cnt = ;
REP(i,,*n) vis[i] = ;
REP(i,,m) {
scanf("%d%d", u+i, v+i);
if (!vis[u[i]]&&!vis[v[i]]&&cnt<n) {
a[i]=vis[u[i]]=vis[v[i]]=;
++cnt;
}
else a[i] = ;
}
if (cnt==n) {
puts("Matching");
REP(i,,m) if (a[i]) printf("%d ",i);
}
else {
puts("IndSet");
cnt = ;
REP(i,,*n) if (!vis[i]) {
printf("%d ",i);
if (++cnt==n) break;
}
}
puts("");
} int main() {
int t;
scanf("%d", &t);
while (t--) work();
}
1198 D Rectangle Painting 1
大意: 给定$n^2$棋盘, 每个格子黑或白, 每次操作选择一个$w\times h$的矩形染成白色, 花费为$max(w,h)$, 求最少花费使得棋盘全白.
范围很小, 直接暴力区间$dp$吧.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head const int N = ;
int n;
char s[N];
int sum[N][N], dp[N][N][N][N]; int get(int a, int b, int c, int d) {
--a,--b;
return sum[c][d]-sum[a][d]-sum[c][b]+sum[a][b];
}
void chkmin(int &x, int y) {x>y?x=y:;}
int main() {
scanf("%d", &n);
REP(i,,n) {
scanf("%s", s+);
REP(j,,n) sum[i][j]=sum[i-][j]+sum[i][j-]-sum[i-][j-]+(s[j]=='#');
}
REP(j,,n) PER(i,,j) REP(b,,n) PER(a,,b) if (get(i,a,j,b)) {
int &r = dp[i][a][j][b] = max(b-a+,j-i+);
REP(x,a,b-) {
chkmin(r,dp[i][a][j][x]+dp[i][x+][j][b]);
}
REP(x,i,j-) {
chkmin(r,dp[i][a][x][b]+dp[x+][a][j][b]);
}
}
printf("%d\n",dp[][][n][n]);
}
1198 E Rectangle Painting 2
大意: $1198D$的花费改为$min(w,h)$.
因为花费是取$min$, 那么相当于每次可以选一行或一列染成白色, 所以可以离散化以后转化为二分图最小带权覆盖问题. 可以参考poj2226
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std; const int N = 1e6+, S = N-, T = N-, INF = 0x3f3f3f3f;
int n, m;
int vx[N], vy[N];
struct {int x1,y1,x2,y2;} a[N];
struct edge {
int to,w,next;
edge(int to=,int w=,int next=):to(to),w(w),next(next){}
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=;
queue<int> Q;
int bfs() {
REP(i,,*vx+*vy) dep[i]=INF,vis[i]=,cur[i]=head[i];
dep[S]=INF,vis[S]=,cur[S]=head[S];
dep[T]=INF,vis[T]=,cur[T]=head[T];
dep[S]=,Q.push(S);
while (Q.size()) {
int u = Q.front(); Q.pop();
for (int i=head[u]; i; i=e[i].next) {
if (dep[e[i].to]>dep[u]+&&e[i].w) {
dep[e[i].to]=dep[u]+;
Q.push(e[i].to);
}
}
}
return dep[T]!=INF;
}
int dfs(int x, int w) {
if (x==T) return w;
int used = ;
for (int i=cur[x]; i; i=e[i].next) {
cur[x] = i;
if (dep[e[i].to]==dep[x]+&&e[i].w) {
int f = dfs(e[i].to,min(w-used,e[i].w));
if (f) used+=f,e[i].w-=f,e[i^].w+=f;
if (used==w) break;
}
}
return used;
}
int dinic() {
int ans = ;
while (bfs()) ans+=dfs(S,INF);
return ans;
}
void add(int u, int v, int w) {
e[++cnt] = edge(v,w,head[u]);
head[u] = cnt;
e[++cnt] = edge(u,,head[v]);
head[v] = cnt;
} int main() {
int n, m;
scanf("%d%d", &n, &m);
if (!m) return puts(""),;
REP(i,,m) {
scanf("%d%d%d%d", &a[i].x1, &a[i].y1, &a[i].x2, &a[i].y2);
vx[++*vx] = a[i].x1;
vx[++*vx] = ++a[i].x2;
vy[++*vy] = a[i].y1;
vy[++*vy] = ++a[i].y2;
}
sort(vx+,vx++*vx);
sort(vy+,vy++*vy);
*vx = unique(vx+,vx++*vx)-vx-;
*vy = unique(vy+,vy++*vy)-vy-;
REP(i,,m) {
a[i].x1 = lower_bound(vx+,vx++*vx,a[i].x1)-vx;
a[i].x2 = lower_bound(vx+,vx++*vx,a[i].x2)-vx-;
a[i].y1 = lower_bound(vy+,vy++*vy,a[i].y1)-vy;
a[i].y2 = lower_bound(vy+,vy++*vy,a[i].y2)-vy-;
REP(x,a[i].x1,a[i].x2) REP(y,a[i].y1,a[i].y2) {
add(x,y+*vx,INF);
}
}
REP(i,,*vx-) add(S,i,vx[i+]-vx[i]);
REP(i,,*vy-) add(*vx+i,T,vy[i+]-vy[i]);
printf("%d\n", dinic());
}
1198 F GCD Groups 2
大意: 给定$n$个数, 求划分为两个集合, 使得每个集合所有元素的$gcd$相同.
