poj 2987 Firing【最大闭合子图】

来源：互联网发布：云端软件平台1.0 编辑：程序博客网时间：2024/05/21 23:00
题目链接：http://poj.org/problem?id=2987
最大闭合子图资料：http://blog.sina.com.cn/s/blog_48f85e1d0100mxem.html
最大闭合子图权值为：所有点值大于0之和 - 最大流
闭合子图点为s能到达的点。
代码：
#include <iostream>#include <algorithm>#include <set>#include <map>#include <string.h>#include <queue>#include <sstream>#include <stdio.h>#include <math.h>#include <stdlib.h>#include <string>using namespace std;const long long MAXN = 50000;//点数的最大值const long long MAXM = 610000;//边数的最大值const long long INF = 0x3f3f3f3f;struct Edge{    long long to, next;    long long cap, flow;}edge[MAXM];//注意是MAXMlong long tol;long long head[MAXN];long long gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];void init(){    tol = 0;    memset(head, -1, sizeof(head));}//加边，单向图三个参数，双向图四个参数void addedge(long long u, long long v, long long w, long long rw = 0){    edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u];    edge[tol].flow = 0; head[u] = tol++;    edge[tol].to = u; edge[tol].cap = rw; edge[tol].next = head[v];    edge[tol].flow = 0; head[v] = tol++;}//输入参数：起点、终点、点的总数//点的编号没有影响，只要输入点的总数int Q[MAXN];void BFS(long long start,long long end){    memset(dep, -1, sizeof(dep));    memset(gap, 0, sizeof(gap));    gap[0] = 1;    long long front = 0, rear = 0;    dep[end] = 0;    Q[rear++] = end;    while (front != rear)    {        long long 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]]++;        }    }}long long S[MAXN];long long sap(long long start, long long end, long long N){    BFS(start, end);    memcpy(cur, head, sizeof(head));    long long u = start;    long long top = 0;    long long  ans = 0;    while (dep[start] < N)    {        if (u == end)        {            long long Min = INF;            int inser;            for (int 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 (int 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;        long long v;        for (long long 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;        }        long long Min = N;        for (int 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;}long long n,m;long long a[100010];long long num;long long vis[10010];void dfs(int x){    vis[x]=1;    if (x!=0)       num++;    for(int i=head[x];i!=-1;i=edge[i].next)    {        if (edge[i].cap-edge[i].flow>0 && !vis[edge[i].to])            dfs(edge[i].to);    }}int main(){    while (scanf("%lld%lld",&n,&m)!=EOF)    {        init();        long long sum = 0;        num=0;        memset(vis,0,sizeof(vis));        for(long long i=1;i<=n;i++)        {            scanf("%lld",&a[i]);            if (a[i] >0)            {                addedge(0,i,a[i]);                sum+=a[i];            }            else            {                addedge(i,n+1,-a[i]);            }        }        long long a,b;        while (m--)        {            scanf("%lld%lld",&a,&b);            addedge(a,b,1e8);        }        long long ans = sap(0,n+1,n+2);        dfs(0);        //for(int i=0;i<num;i++)        //   printf("%d ",path[i]);        printf("%lld %lld\n",num,sum - ans);    }    return 0;}
0 0