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+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int u[N], v[N], a[N], vis[N]; void work() { int n, m; scanf("%d%d", &n, &m); int cnt = 0; REP(i,1,3*n) vis[i] = 0; REP(i,1,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]]=1; ++cnt; } else a[i] = 0; } if (cnt==n) { puts("Matching"); REP(i,1,m) if (a[i]) printf("%d ",i); } else { puts("IndSet"); cnt = 0; REP(i,1,3*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+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 55; 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:0;} int main() { scanf("%d", &n); REP(i,1,n) { scanf("%s", s+1); REP(j,1,n) sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+(s[j]=='#'); } REP(j,1,n) PER(i,1,j) REP(b,1,n) PER(a,1,b) if (get(i,a,j,b)) { int &r = dp[i][a][j][b] = max(b-a+1,j-i+1); REP(x,a,b-1) { chkmin(r,dp[i][a][j][x]+dp[i][x+1][j][b]); } REP(x,i,j-1) { chkmin(r,dp[i][a][x][b]+dp[x+1][a][j][b]); } } printf("%d\n",dp[1][1][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+10, S = N-2, T = N-1, 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=0,int w=0,int next=0):to(to),w(w),next(next){} } e[N]; int head[N], dep[N], vis[N], cur[N], cnt=1; queue<int> Q; int bfs() { REP(i,1,*vx+*vy) dep[i]=INF,vis[i]=0,cur[i]=head[i]; dep[S]=INF,vis[S]=0,cur[S]=head[S]; dep[T]=INF,vis[T]=0,cur[T]=head[T]; dep[S]=0,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]+1&&e[i].w) { dep[e[i].to]=dep[u]+1; Q.push(e[i].to); } } } return dep[T]!=INF; } int dfs(int x, int w) { if (x==T) return w; int used = 0; for (int i=cur[x]; i; i=e[i].next) { cur[x] = i; if (dep[e[i].to]==dep[x]+1&&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^1].w+=f; if (used==w) break; } } return used; } int dinic() { int ans = 0; 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,0,head[v]); head[v] = cnt; } int main() { int n, m; scanf("%d%d", &n, &m); if (!m) return puts("0"),0; REP(i,1,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+1,vx+1+*vx); sort(vy+1,vy+1+*vy); *vx = unique(vx+1,vx+1+*vx)-vx-1; *vy = unique(vy+1,vy+1+*vy)-vy-1; REP(i,1,m) { a[i].x1 = lower_bound(vx+1,vx+1+*vx,a[i].x1)-vx; a[i].x2 = lower_bound(vx+1,vx+1+*vx,a[i].x2)-vx-1; a[i].y1 = lower_bound(vy+1,vy+1+*vy,a[i].y1)-vy; a[i].y2 = lower_bound(vy+1,vy+1+*vy,a[i].y2)-vy-1; REP(x,a[i].x1,a[i].x2) REP(y,a[i].y1,a[i].y2) { add(x,y+*vx,INF); } } REP(i,1,*vx-1) add(S,i,vx[i+1]-vx[i]); REP(i,1,*vy-1) add(*vx+i,T,vy[i+1]-vy[i]); printf("%d\n", dinic()); }
1198 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.
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