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线性规划与网络流24题 02太空飞行计划问题

2011-10-03 10:25 495 查看
这个题的shut5.in的数据会是多解,所以评测可能出错。。。。。。很好的一个题。。。重点在向最小割的转化。。。。。建议看一下胡泊涛的论文《最小割模型在信息学竞赛中的应用》 里面很详细。。也很好。。。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define inf 1<<30
#define M 100000
#define N 10000
#define cc(m,v) memset(m,v,sizeof(m))

struct node {
int u, v, f, next;
} edge[M];
int head
, p, lev
, cur
;
int que[M];

void ainit() {
p = 0, cc(head, -1);
}

bool bfs(int s, int t) {
int i, u, v, qin = 0, qout = 0;
cc(lev, -1), lev[s] = 0, que[qin++] = s;
while (qout != qin) {
u = que[qout++];
for (i = head[u]; i != -1; i = edge[i].next)
if (edge[i].f > 0 && lev[v = edge[i].v] == -1) {
lev[v] = lev[u] + 1, que[qin++] = v;
if (v == t) return 1;
}
}
return 0;
}

int dinic(int s, int t) {
int i, f, k, u, qin;
int flow = 0;
while (bfs(s, t)) {
memcpy(cur, head, sizeof (head));
u = s, qin = 0;
while (1) {
if (u == t) {
for (k = 0, f = inf; k < qin; k++)
if (edge[que[k]].f < f) f = edge[que[i = k]].f;
for (k = 0; k < qin; k++)
edge[que[k]].f -= f, edge[que[k]^1].f += f;
flow += f, u = edge[que[qin = i]].u;
}
for (i = cur[u]; cur[u] != -1; i = cur[u] = edge[cur[u]].next)
if (edge[i].f > 0 && lev[u] + 1 == lev[edge[i].v]) break;
if (cur[u] != -1)
que[qin++] = cur[u], u = edge[cur[u]].v;
else {
if (qin == 0) break;
lev[u] = -1, u = edge[que[--qin]].u;
}
}
}
return flow;
}

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

int main() {
int n, m, i, cost, u, ans, sum;
while (scanf("%d%d", &m, &n) != -1) {
ainit();
for (i = 1, sum = 0; i <= m; i++) {
scanf("%d", &cost);
sum += cost, addedge(0, i, cost);
while (getchar() != '\n') {
scanf("%d", &u);
addedge(i, u + m, inf);
}
}
for (i = 1; i <= n; i++) {
scanf("%d", &u);
addedge(i + m, n + m + 1, u);
}
ans = dinic(0, n + m + 1);
for (i = 1; i <= m; i++) if (lev[i] != -1)
printf("%d ", i);
printf("\n");
for (i = m + 1; i <= n + m; i++) if (lev[i] != -1)
printf("%d ", i - m);
printf("\n%d\n",sum - ans);
}
return 0;
}
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