HDU 3126 Nova [2009 Asia Wuhan Regional Contest Online]

题目大意

有n个巫妖，m个精灵，k棵树，他们都有自己的坐标表示自己的位置。巫妖有攻击范围和冷却时间，树有覆盖范围。

如果某个巫妖攻击精灵的路线（他俩之间的连线）经过树的覆盖范围，表示精灵被树挡住巫妖攻击不到。求巫妖杀死所有精灵的时间，若无法全部杀死输出-1；

解题思路：

判断巫妖是否能打到精灵用线段与点的最短距离来判断，若最短距离小于树的覆盖范围，就攻击不到。

最小时间可以跑费用流来解决，也可以二分图的最优匹配。

下面是代码：

#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <cctype>#include <algorithm>#define eps 1e-6#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define zero(a) fabs(a)<eps#define iabs(x)  ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )using namespace std;struct Point{    double x,y;} hitted[205];struct node1{    double r;    int t;    Point p;} att[205];struct node3{    double r;    Point p;} tree[205];int n,m,K;double Distance(Point a, Point b){    return sqrt((a.x-b.x)*(a.x-b.x)*1.0 + 1.0*(a.y-b.y)*(a.y-b.y));}double xmult(Point p1, Point p2, Point p){    return (p1.x-p.x)*(p2.y-p.y) - (p2.x-p.x)*(p1.y-p.y);}double disptoseg(Point p, Point a ,Point b){    Point t = p;    t.x += a.y-b.y, t.y += b.x-a.x;    if(xmult(a,t,p)*xmult(b,t,p)>eps)        return Distance(p,a) < Distance(p,b)? Distance(p,a) : Distance(p,b);    return fabs(xmult(p, a, b))/Distance(a, b);}int head[450],cnt;struct node{    int u,v,w,f,next;} edge[230*230*20];void addedge(int u,int v,int w,int f){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].f=f;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].u=v;    edge[cnt].v=u;    edge[cnt].w=-w;    edge[cnt].f=0;    edge[cnt].next=head[v];    head[v]=cnt++;}int pre[450];bool vis[450];bool spfa(){    int i;    //clearall(pre,-1);    pre[n+m+1]=-1;    pre[n+m]=-1;    int dis[450];    clearall(dis,0x3f3f);    queue <int >q;    dis[n+m]=0;    vis[n+m]=true;    q.push(n+m);    while(!q.empty())    {        int t=q.front();        q.pop();        i=head[t];        vis[t]=false;        while(i!=-1)        {            if(edge[i].f>0&&dis[edge[i].v]>dis[t]+edge[i].w)            {                dis[edge[i].v]=dis[t]+edge[i].w;                pre[edge[i].v]=i;                if(!vis[edge[i].v])                {                    vis[edge[i].v]=true;                    q.push(edge[i].v);                }            }            i=edge[i].next;        }    }    if(pre[n+m+1]==-1)return false;    return true;}bool cnthit[205];double input(){    char s=getchar();    double ans=0;    while((s<'0'||s>'9')&&s!='-')    {        s=getchar();    }    bool flat=false;    if(s=='-')    {        flat=true;        s=getchar();    }    while(s>='0'&&s<='9')    {        ans*=10;        ans+=s-'0';        s=getchar();    }    if(flat)ans=-ans;    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&K);        for(int i=0; i<n; i++)        {            att[i].p.x=input();            att[i].p.y=input();            att[i].r=input();            att[i].t=input();        }        for(int i=0; i<m; i++)        {            hitted[i].x=input();            hitted[i].y=input();        }        for(int i=0; i<K; i++)        {            tree[i].p.x=input();            tree[i].p.y=input();            tree[i].r=input();        }        clearall(head,-1);        clearall(cnthit,0);        cnt=0;        for(int i=0; i<n; i++)        {            int cnt1=0;            for(int j=0; j<m; j++)            {                bool flat=true;                if(Distance(att[i].p, hitted[j]) <= att[i].r)                {                    for(int k=0; k<K; k++)                    {                        if(disptoseg(tree[k].p,att[i].p,hitted[j])<=tree[k].r)                        {                            flat=false;                            break;                        }                    }                    if(flat)                    {                        cnthit[j]=true;                        addedge(i,n+j,0,1);                        addedge(n+m,i,att[i].t*cnt1,1);                        cnt1++;                    }                }            }        }        for(K=0; K<m; K++)        {            if(cnthit[K]==false)            {                break;            }        }        if(K!=m)        {            puts("-1");            continue;        }        for(int i=0; i<m; i++)        {            addedge(i+n,n+m+1,0,1);        }        int ans=0;        while(spfa())        {            int max1=inf;            int p=pre[n+m+1];            while(p!=-1)            {                max1=min(max1,edge[p].f);                p=pre[edge[p].u];            }            p=pre[n+m+1];            while(p!=-1)            {                edge[p].f-=max1;                edge[p^1].f+=max1;                p=pre[edge[p].u];            }        }        for(int i=head[n+m];i!=-1;i=edge[i].next)        {            if(edge[i].f==0)ans=max(ans,edge[i].w);        }        printf("%d\n",ans);    }    return 0;}