2014-2015 ACM-ICPC, Asia Tokyo Regional Contest C題:Shopping [贪心+并查集]
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题意:给出N+2个点,每个点的编号为0...N+1,每一对点之间有一条长度为两者编号差的边,问走遍所有点且满足一些约束条件下,从0号点到N+1号点的最短路是多少。约束条件有M个,每一个约束条件给出一个X,Y,表示走X点之前必须先走Y点。(X可能对应多个Y,X<Y)
范围:N<=1000,M<=500
解法:首先贪心发现,如果有一对点X,Y,那么必然要先从X点走到Y点,那么这之间的点为何不都直接走呢?然后从Y再走到X点,这时候因为下标递减,所以这些点大部分都会被满足,假设不满足的点是X2,Y2,那么Y2必然大于Y,这时候只要一开始走到Y2,然后再走到X就好啦,而且这样是更优的,因为X2到Y1这段路走的次数减少了。所以只要建立并茶集,对于每对点X,Y,把X至Y这些点都并到一个集合里,并且记录这个集合里的最大值Max和最小值Min,因为要满足约束条件,所以再往回走一遍,多走的路程为2*(Max-Min),累加起来,再加上N+1就是答案啦。(N+1是原本就要走的路程...)
代码:
#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<iostream>#include<stdlib.h>#include<set>#include<map>#include<queue>#include<vector>#include<bitset>#pragma comment(linker, "/STACK:1024000000,1024000000")template <class T>bool scanff(T &ret){ //Faster Input char c; int sgn; T bit=0.1; if(c=getchar(),c==EOF) return 0; while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar(); sgn=(c=='-')?-1:1; ret=(c=='-')?0:(c-'0'); while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0'); if(c==' '||c=='\n'){ ret*=sgn; return 1; } while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10; ret*=sgn; return 1;}#define inf 1073741823#define llinf 4611686018427387903LL#define PI acos(-1.0)#define lth (th<<1)#define rth (th<<1|1)#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)#define mem(x,val) memset(x,val,sizeof(x))#define mkp(a,b) make_pair(a,b)#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define pb(x) push_back(x)using namespace std;typedef long long ll;typedef pair<int,int> pii;#define NN 100100int a[NN],n,m;int f[NN],minv[NN],maxv[NN];int find(int x){ while(x!=f[x]){ f[x]=f[f[x]]; x=f[x]; } return x;}void connec(int x,int y){ int i=find(x); int j=find(y); if(i==j)return; minv[j]=min(minv[i],minv[j]); maxv[j]=max(maxv[i],maxv[j]); f[i]=j;}bool vis[NN];int main(){ scanff(n);scanff(m); rep(i,1,n){ f[i]=i; minv[i]=i; maxv[i]=i; } rep(i,1,m){ int l,r; scanff(l); scanff(r); rep(i,l,r-1)connec(i,i+1); } int ans=n+1; rep(i,1,n)vis[find(i)]=1; rep(i,1,n){ if(vis[i])ans+=2*(maxv[i]-minv[i]); } printf("%d\n",ans); return 0;}
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