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 MB
Submit: 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

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