您的位置:首页 > 其它

【BZOJ1305】[CQOI2009]dance跳舞【最大流】【二分】

2016-04-25 22:24 495 查看
【题目链接】

看了【POPOQQQ的题解】,建图比较厉害。

写网络流第一次把点开小了...orz

/* Pigonometry */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 255, maxm = 10005, maxg = 55, maxq = 10000, inf = 0x3f3f3f3f;

int n, m, head[maxn], cur[maxn], cnt, A[maxg][maxg], depth[maxn], bg, ed, q[maxq];

struct _edge {
int v, w, next;
} g[maxm << 1];

inline void add(int u, int v, int w) {
g[cnt] = (_edge){v, w, head[u]};
head[u] = cnt++;
}

inline void insert(int u, int v, int w) {
add(u, v, w); add(v, u, 0);
}

inline bool bfs() {
for(int i = 0; i <= ed; i++) depth[i] = -1;
int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1;
while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) {
depth[g[i].v] = depth[u] + 1;
if(g[i].v == ed) return 1;
q[t++] = g[i].v;
}
return 0;
}

inline int dfs(int x, int flow) {
if(x == ed) return flow;
int left = flow;
for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) {
int tmp = dfs(g[i].v, min(left, g[i].w));
left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp;
if(g[i].w) cur[x] = i;
if(!left) return flow;
}
if(left == flow) depth[x] = -1;
return flow - left;
}

inline bool check(int mid) {
for(int i = 0; i <= ed; i++) head[i] = -1; cnt = 0;

for(int i = 1; i <= n; i++) {
insert(bg, i, mid);
insert(n + n + n + i, ed, mid);
insert(i, n + i, m);
insert(n + n + i, n + n + n + i, m);
}
for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++)
if(A[i][j]) insert(i, n + n + n + j, 1);
else insert(n + i, n + n + j, 1);

int ans = 0;
while(bfs()) {
for(int i = 0; i <= ed; i++) cur[i] = head[i];
ans += dfs(bg, inf);
}

return n * mid == ans;
}

char str[maxg];

int main() {
scanf("%d%d", &n, &m); bg = 0; ed = (n << 2) + 1;
for(int i = 1; i <= n; i++) {
scanf("%s", str + 1);
for(int j = 1; j <= n; j++) A[i][j] = (str[j] == 'Y');
}

int l = 0, r = n;
while(l <= r) {
int mid = l + r >> 1;
if(check(mid)) l = mid + 1;
else r = mid - 1;
}

printf("%d\n", r);
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: