poj 1984

带权并查集

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int f[40010];int s1[40010],s2[40010];  //s1为水平，s2为竖直int ans[10010];struct Road{    int u,v,c;    char dir;}road[40010];struct Query{    int u,v,id;    int num;}query[10010];void init(int n){    for(int i=1;i<=n;i++){        f[i]=i;        s1[i]=s2[i]=0;    }}bool cmp(Query a,Query b){    return a.id<b.id;}int find(int u){    if(f[u]!=u){        int tem=f[u];        f[u]=find(tem);        s1[u]+=s1[tem];        s2[u]+=s2[tem];    }    return f[u];}void Union(int i){    int u=find(road[i].u);    int v=find(road[i].v);    if(u!=v){        f[u]=v;        if(road[i].dir=='E'){            s1[u]=-s1[road[i].u]+road[i].c+s1[road[i].v];            s2[u]=-s2[road[i].u]+s2[road[i].v];        }        else if(road[i].dir=='W'){            s1[u]=-s1[road[i].u]-road[i].c+s1[road[i].v];            s2[u]=-s2[road[i].u]+s2[road[i].v];        }        else if(road[i].dir=='N'){            s1[u]=-s1[road[i].u]+s1[road[i].v];            s2[u]=-s2[road[i].u]+road[i].c+s2[road[i].v];        }        else if(road[i].dir=='S'){            s1[u]=-s1[road[i].u]+s1[road[i].v];            s2[u]=-s2[road[i].u]-road[i].c+s2[road[i].v];        }    }}int main(){    int n,m,i,j,k;    int now,u,v;    char dir;    scanf("%d %d",&n,&m);    init(n);    for(i=1;i<=m;i++)        scanf("%d %d %d %c",&road[i].u,&road[i].v,&road[i].c,&road[i].dir);    scanf("%d",&k);    for(i=1;i<=k;i++){        scanf("%d %d %d",&query[i].u,&query[i].v,&query[i].id);        query[i].num=i;    }    sort(query+1,query+1+k,cmp);    now=1;    for(i=1;i<=m;i++){        Union(i);        while(query[now].id==i && now<=k){            u=find(query[now].u);            v=find(query[now].v);            if(u==v)                ans[query[now].num]=abs(s1[query[now].u]-s1[query[now].v])+abs(s2[query[now].u]-s2[query[now].v]);            else                ans[query[now].num]=-1;            now++;        }        if(now==k+1)            break;    }    for(i=1;i<=k;i++)        printf("%d\n",ans[i]);    return 0;}