【网络流】 HDU 4183 Pahom on Water 拆点

来源：互联网发布：各种协议端口号编辑：程序博客网时间：2024/06/17 02:07
题意：求两条路能从 400.0 -> 789.0 且这两条路不想交（除了端点400，789 ）
求只能走一次的网络流需要用到拆点，
将点i 拆成 i 和 i+n i->i+n的容量为经过的次数（这题为1 ）
若i 能到达 j 则连接 i+n-> j
#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <iostream>#include <algorithm>#include <sstream>#include <cmath>using namespace std;#include <queue>#include <stack>#include <vector>#include <deque>#include <set>#include <map>#include <time.h>;#define cler(arr, val)    memset(arr, val, sizeof(arr))#define FOR(i,a,b)  for(int i=a;i<=b;i++)#define IN   freopen ("in.txt" , "r" , stdin);#define OUT  freopen ("out.txt" , "w" , stdout);typedef long long  LL;const int MAXN = 514;const int MAXM = 40101;const int INF = 0x3f3f3f3f;const int mod = 1000000007;struct Edge{    int to,next,cap,flow;} edge[MAXM]; //注意是MAXMint tol;int head[MAXN];int gap[MAXN],dep[MAXN],cur[MAXN];void init(){    tol = 0;    memset(head,-1,sizeof (head));}void addedge (int u,int v,int w,int rw = 0){    edge[tol].to = v;    edge[tol].cap = w;    edge[tol].flow = 0;    edge[tol].next = head[u];    head[u] = tol++;    edge[tol].to = u;    edge[tol].cap = rw;    edge[tol].flow = 0;    edge[tol].next = head[v];    head[v] = tol++;}int Q[MAXN];void BFS(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0] = 1;    int front = 0, rear = 0;    dep[end] = 0;    Q[rear++] = end;    while(front != rear)    {        int u = Q[front++];        for(int i = head[u]; i !=  -1; i = edge[i].next)        {            int v = edge[i]. to;            if(dep[v] != -1)continue;            Q[rear++] = v;            dep[v] = dep[u] + 1;            gap[dep[v]]++;        }    }}int S[MAXN];int sap(int start,int end, int N){    BFS(start,end);    memcpy(cur,head,sizeof(head));    int top = 0;    int u = start;    int ans = 0;    int i;    while(dep[start] < N)    {        if(u == end)        {            int Min = INF;            int inser;            for( i = 0; i < top; i++)            {                if(Min > edge[S[i]].cap - edge[S[i]].flow)                {                    Min = edge[S[i]].cap - edge[S[i]].flow;                    inser = i;                }            }            for( i = 0; i < top; i++)            {                edge[S[i]]. flow += Min;                edge[S[i]^1].flow -= Min;            }            ans += Min;            top = inser;            u = edge[S[top]^1].to;            continue;        }        bool flag =  false;        int v;        for( i = cur[u]; i != -1; i = edge[i]. next)        {            v = edge[i]. to;            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])            {                flag =  true;                cur[u] = i;                break;            }        }        if(flag)        {            S[top++] = cur[u];            u = v;            continue;        }        int Min = N;        for( i = head[u]; i !=  -1; i = edge[i].next)        {            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)            {                Min = dep[edge[i].to];                cur[u] = i;            }        }        gap[dep[u]]--;        if(!gap[dep[u]]) return ans;        dep[u] = Min + 1;        gap[dep[u]]++;        if(u != start)u = edge[S[--top]^1].to;    }    return ans;}double gao(double x,double y,double a,double b){    return sqrt((x-a)*(x-a)+(y-b)*(y-b));}struct point{    double f,x,y,r;    point(){};}a[202];bool cmp(point a,point b){    return a.f<b.f;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif    int t,n;    scanf("%d",&t);    while(t--)    {        init();        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%lf%lf%lf%lf",&a[i].f,&a[i].x,&a[i].y,&a[i].r);        sort(a,a+n,cmp);        addedge(0,n,1);        addedge(n-1,2*n-1,1);        for(int i=0;i<n;i++)        {            addedge(i,i+n,1);            for(int j=i+1;j<n;j++)            {                if(a[i].f<a[j].f&&gao(a[i].x,a[i].y,a[j].x,a[j].y)<(a[i].r+a[j].r))                    addedge(i+n,j,1);            }        }        if(sap(0,2*n-1,2*n)==2)            puts("Game is VALID");        else puts("Game is NOT VALID");    }    return 0;}
0 0