BZOJ 1565: [NOI2009]植物大战僵尸( 最小割 )
先拓扑排序搞出合法的, 然后就是最大权闭合图模型了....
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#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define Id(x, y) ((x) * c + (y))
const int maxv = 700;
const int INF = 10000000;
int r, c;
int deg[maxv], w[maxv], q[maxv], Id[maxv];
struct edge {
int t;
edge* nxt;
} E[maxv * maxv], *pt = E, *Head[maxv];
inline void AddEdge(int u, int v) {
deg[pt->t = v]++, pt->nxt = Head[u], Head[u] = pt++;
}
namespace G {
struct edge {
int t, c;
edge *nxt, *rev;
} E[maxv * maxv], *e = E, *Head[maxv], *p[maxv], *cur[maxv];
int V, h[maxv], cnt[maxv];
inline void Add(int u, int v, int c) {
e->t = v, e->c = c, e->nxt = Head[u], Head[u] = e++;
}
inline void AddEdge(int u, int v, int c) {
Add(u, v, c), Add(v, u, 0);
Head[u]->rev = Head[v];
Head[v]->rev = Head[u];
}
void Work() {
int S = V++, T = V++, ans = 0;
for(int i = r * c; i--; ) if(~Id[i]) {
if(w[i] > 0) {
ans += w[i], AddEdge(S, Id[i], w[i]);
} else
AddEdge(Id[i], T, -w[i]);
}
for(int i = 0; i < V; i++)
cnt[i] = h[i] = 0, cur[i] = Head[i];
cnt[0] = V;
for(int x = S, A = INF; h[S] < V; ) {
for(e = cur[x]; e; e = e->nxt)
if(e->c && h[e->t] + 1 == h[x]) break;
if(e) {
p[e->t] = cur[x] = e;
A = min(A, e->c);
if((x = e->t) == T) {
for(; x != S; x = p[x]->rev->t)
p[x]->c -= A, p[x]->rev->c += A;
ans -= A, A = INF;
}
} else {
if(!--cnt[h[x]]) break;
h[x] = V;
for(e = Head[x]; e; e = e->nxt) if(h[e->t] + 1 < h[x] && e->c)
h[x] = h[e->t] + 1, cur[x] = e;
cnt[h[x]]++;
if(x != S) x = p[x]->rev->t;
}
}
printf("%d\n", ans);
}
}
void Init() {
int n;
scanf("%d%d", &r, &c);
memset(deg, 0, sizeof deg);
memset(Id, -1, sizeof Id);
for(int i = 0; i < r; i++)
for(int j = 0; j < c; j++) {
scanf("%d%d", w + Id(i, j), &n);
while(n--) {
int x, y; scanf("%d%d", &x, &y);
AddEdge(Id(i, j), Id(x, y));
}
if(j) AddEdge(Id(i, j), Id(i, j - 1));
}
}
void BFS() {
int h = 0, t = 0, x, &V = G::V = 0;
for(int i = r * c; i--; )
if(!deg[i]) q[t++] = i;
while(h < t) {
Id[x = q[h++]] = V++;
for(edge* e = Head[x]; e; e = e->nxt)
if(!--deg[e->t]) q[t++] = e->t;
}
}
void Work() {
for(int i = r * c; i--; ) if(~Id[i])
for(edge* e = Head[i]; e; e = e->nxt)
if(~Id[e->t]) G::AddEdge(Id[e->t], Id[i], INF);
G::Work();
}
int main() {
Init();
BFS();
Work();
return 0;
}
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1565: [NOI2009]植物大战僵尸
Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 1807 Solved: 835
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Description
Input
Output
仅包含一个整数，表示可以获得的最大能源收入。注意，你也可以选择不进行任何攻击，这样能源收入为0。
Sample Input
3 2
10 0
20 0
-10 0
-5 1 0 0
100 1 2 1
100 0
10 0
20 0
-10 0
-5 1 0 0
100 1 2 1
100 0
Sample Output
25
HINT
在样例中, 植物P1,1可以攻击位置(0,0), P2, 0可以攻击位置(2,1)。
一个方案为，首先进攻P1,1, P0,1，此时可以攻击P0,0 。共得到能源收益为(-5)+20+10 = 25。注意, 位置(2,1)被植物P2,0保护，所以无法攻击第2行中的任何植物。
【大致数据规模】
约20%的数据满足1 ≤ N, M ≤ 5；
约40%的数据满足1 ≤ N, M ≤ 10；
约100%的数据满足1 ≤ N ≤ 20，1 ≤ M ≤ 30，-10000 ≤ Score ≤ 10000 。
Source
