Destroying The Graph | Time Limit: 2000MS | | Memory Limit: 65536K | | Total 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 6
1 2 3
4 2 1
1 2
1 1
3 2
1 2
3 1
2 3
Sample Output 5
3
1 +
2 -
2 + Source Northeastern Europe 2003, Northern Subregion 一道典型的最小點(diǎn)權(quán)覆蓋問(wèn)題，SAP速度很好看，之后需要搜索一下用過(guò)的點(diǎn)，輸出即可。 代碼： #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) //不同于單純的SAP，加入了一個(gè)搜素點(diǎn)集元素的函數(shù)，通過(guò)head數(shù)組的記錄信息搜索。
{
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;
} |