POJ2125:Destroying The Graph



Destroying The Graph

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 4718 Accepted: 1436 Special Judge

Description

Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to destroy it. In a move he may take any vertex of the graph and remove either all arcs incoming into this vertex, or all arcs outgoing from this vertex. 
Alice assigns two costs to each vertex: Wi⁺ and Wi^-. If Bob removes all arcs incoming into the i-th vertex he pays Wi⁺ dollars to Alice, and if he removes outgoing arcs he pays Wi^- dollars. 
Find out what minimal sum Bob needs to remove all arcs from the graph.

Input

Input file describes the graph Alice has drawn. The first line of the input file contains N and M (1 <= N <= 100, 1 <= M <= 5000). The second line contains N integer numbers specifying Wi⁺. The third line defines Wi^- in a similar way. All costs are positive and do not exceed 10⁶ . Each of the following M lines contains two integers describing the corresponding arc of the graph. Graph may contain loops and parallel arcs.

Output

On the first line of the output file print W --- the minimal sum Bob must have to remove all arcs from the graph. On the second line print K --- the number of moves Bob needs to do it. After that print K lines that describe Bob's moves. Each line must first contain the number of the vertex and then '+' or '-' character, separated by one space. Character '+' means that Bob removes all arcs incoming into the specified vertex and '-' that Bob removes all arcs outgoing from the specified vertex.

Sample Input

3 61 2 34 2 11 21 13 21 23 12 3

Sample Output

531 +2 -2 +

Source

Northeastern Europe 2003, Northern Subregion



一道典型的最小点权覆盖问题，SAP速度很好看，之后需要搜索一下用过的点，输出即可。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define N 10010
#define M 20010
#define inf 1 << 30
#define eps 1 << 29
using namespace std;

int mark[N];
int cnt, n, m, s, t;
int head[N];
int NN;

struct edge
{
    int v, next, w;
} edge[M];

void addedge(int u, int v, int w)
{
    edge[cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
    edge[cnt].v = u;
    edge[cnt].w = 0;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}

int sap()
{
    int pre[N], cur[N], dis[N], gap[N];
    int flow = 0, aug = inf, u;
    bool flag;
    for (int i = 1; i <= NN; ++i)
    {
        cur[i] = head[i];
        gap[i] = dis[i] = 0;
    }
    gap[s] = NN;
    u = pre[s] = s;
    while (dis[s] < NN)
    {
        flag = 0;
        for (int &j = cur[u]; j != -1; j = edge[j].next)
        {
            int v = edge[j].v;
            if (edge[j].w > 0 && dis[u] == dis[v] + 1)
            {
                flag = 1;
                if (edge[j].w < aug) aug = edge[j].w;
                pre[v] = u;
                u = v;
                if (u == t)
                {
                    flow += aug;
                    while (u != s)
                    {
                        u = pre[u];
                        edge[cur[u]].w -= aug;
                        edge[cur[u] ^ 1].w += aug;
                    }
                    aug = inf;
                }
                break;
            }
        }
        if (flag)
            continue;
        int mindis = NN;
        for (int j = head[u]; j != -1; j = edge[j].next)
        {
            int v = edge[j].v;
            if (edge[j].w > 0 && dis[v] < mindis)
            {
                mindis = dis[v];
                cur[u] = j;
            }
        }
        if ((--gap[dis[u]]) == 0)
            break;
        gap[dis[u] = mindis + 1]++;
        u = pre[u];
    }
    return flow;
}

void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(mark, 0, sizeof(mark));
}

void dfs(int x)      //不同于单纯的SAP，加入了一个搜素点集元素的函数，通过head数组的记录信息搜索。
{
    mark[x] = 1;
    for (int i = head[x]; i; i = edge[i].next)
    {
        if (edge[i].w > 0 && !mark[edge[i].v]) dfs(edge[i].v);
    }
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        init();
        int wp[101], wm[101], w, len = 1, ans[105];
        s = 0;
        t = 2 * n + 1;
        NN = 2 * n + 2;
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &wp[i]);
            addedge(i + n, t, wp[i]);
        }
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &wm[i]);
            addedge(s, i, wm[i]);
        }
        for (int i = 1; i <= m; ++i)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            addedge(x, y + n, inf);
        };
        w = sap();
        dfs(s);
        for (int i = 1; i <= n; ++i)
        {
            if (!mark[i]) ans[len++] = i;
            if (mark[i + n]) ans[len++] = i + n;
        }
        len--;
        printf("%d\n%d\n", w, len);
        for (int i = 1; i <= len; ++i)
        {
            if (ans[i] <=n) printf("%d -\n", ans[i]);
            else printf("%d +\n", ans[i] - n);
        }
    }
    return 0;
}