hdu3938Portal

来源：互联网发布：物品识别软件编辑：程序博客网时间：2024/06/13 05:04

链接：http://acm.hdu.edu.cn/showproblem.php?pid=3938

题意：给定一个n个点m条边的图，q个询问，每次询问给一个x，求有多少点对u,v使得u->v至少有一条路径满足路径中最大值小于等于x。

分析：其实就是求点对，离线一下，将边和询问都从小到大排序，然后用并查集合并就行了。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=10010;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000007;const int INF=1000000010;const ll MAX=1ll<<55;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;struct node {    int x,y,z;    bool operator < (const node a) const {        return z<a.z;    }}b[5*N],Q[N];int f[N];ll ans[N],g[N];int find_f(int a) {    return f[a]==a ? a:f[a]=find_f(f[a]);}int main(){    int i,k,n,m,q,fa,fb;    ll sum;    while (scanf("%d%d%d", &n, &m, &q)!=EOF) {        for (i=1;i<=m;i++) scanf("%d%d%d", &b[i].x, &b[i].y, &b[i].z);        for (i=1;i<=q;i++) scanf("%d", &Q[i].z),Q[i].x=i;        sort(b+1,b+m+1);        sort(Q+1,Q+q+1);        sum=0;k=1;        for (i=1;i<=n;i++) f[i]=i,g[i]=1;        for (i=1;i<=q;i++) {            for (;k<=m&&b[k].z<=Q[i].z;k++) {                fa=find_f(b[k].x);fb=find_f(b[k].y);                if (fa!=fb) {                    sum+=g[fa]*g[fb];                    if (g[fa]<g[fb]) { f[fa]=fb;g[fb]+=g[fa]; }                    else { f[fb]=fa;g[fa]+=g[fb]; }                }            }            ans[Q[i].x]=sum;        }        for (i=1;i<=q;i++) printf("%I64d\n", ans[i]);    }    return 0;}

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