1143, 3997: Dilworth定理的简单应用
偏序集上的最小链覆盖等价于求最长反链
最小不交链覆盖等于传递闭包后最小链覆盖
最小链覆盖大小等于点数减去二分图最大匹配大小
二分图最小点覆盖大小等于二分图匹配大小
二分图最小点覆盖与二分图最大独立集对偶
建图完后,这个二分图最大独立集等价于最长反链。
于是两道题
1143: [CTSC2008]祭祀river
求偏序集上的最长反链
转换成偏序集上的最小链覆盖
求个闭包,转换成最小路径覆盖,二分图匹配一发
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
const int MAXN = 210;
const int MAXM = MAXN * MAXN * 2;
const int INF = 0x3f3f3f3f;
int head[MAXN], nxt[MAXM], to[MAXM], val[MAXM], tot = 1;
inline void addedge(int b, int e, int v) {
nxt[++tot] = head[b]; to[head[b] = tot] = e; val[tot] = v;
nxt[++tot] = head[e]; to[head[e] = tot] = b; val[tot] = 0;
}
std::queue<int> q;
int S, T, dis[MAXN];
bool bfs() {
for (int i = S; i <= T; ++i) dis[i] = 0;
dis[S] = 1; q.push(S);
while (!q.empty()) {
int t = q.front(); q.pop();
for (int i = head[t]; i; i = nxt[i])
if (val[i] && !dis[to[i]]) {
dis[to[i]] = dis[t] + 1;
q.push(to[i]);
}
}
return dis[T] > 0;
}
int dinic(int u, int minv) {
if (u == T || !minv) return minv;
int t, res = 0;
for (int i = head[u]; i; i = nxt[i])
if (val[i] && dis[to[i]] == dis[u] + 1 && (t = dinic(to[i], std::min(minv, val[i])))) {
val[i] -= t;
val[i ^ 1] += t;
res += t;
minv -= t;
if (!minv) break;
}
if (!res) dis[u] = -1;
return res;
}
int n, m, t1, t2, f[MAXN][MAXN];
int main() {
scanf("%d%d", &n, &m);
S = 0, T = n << 1 | 1;
for (int i = 1; i <= m; ++i) {
scanf("%d%d", &t1, &t2);
f[t1][t2] = true;
}
for (int k = 1; k <= n; ++k)
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
f[i][j] |= f[i][k] & f[k][j];
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
if (f[i][j])
addedge(i, j + n, 1);
for (int i = 1; i <= n; ++i) {
addedge(S, i, 1);
addedge(i + n, T, 1);
}
int ans = n;
while (bfs()) ans -= dinic(S, INF);
printf("%d\n", ans);
return 0;
}
3997: [TJOI2015]组合数学
给出一个网格图,其中某些格子有财宝,每次从左上角出发,只能向下或右走。问至少走多少次才能将财宝捡完。此对此问题变形,假设每个格子中有好多财宝,而每一次经过一个格子至多只能捡走一块财宝,至少走多少次才能把财宝全部捡完。
求偏序集上的最小链覆盖,转换为最长反链覆盖
于是只要DP出反链就好了,前(其实是后)缀和优化一下
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
const int MAXN = 1010;
int f[MAXN][MAXN], suc[MAXN][MAXN], map[MAXN][MAXN], n, m;
inline void getmax(int & x, const int y) { if (x < y) x = y; }
int main() {
int T; scanf("%d", &T);
while (T --> 0) {
scanf("%d%d", &n, &m);
for (int i = 0; i <= n + 1; ++i)
for (int j = 0; j <= m + 1; ++j)
f[i][j] = suc[i][j] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) scanf("%d", map[i] + j);
for (int j = m; j; --j) {
getmax(f[i][j], suc[i - 1][j + 1] + map[i][j]);
suc[i][j] = std::max(suc[i][j + 1], suc[i - 1][j]);
getmax(suc[i][j], f[i][j]);
}
}
printf("%d\n", suc[n][1]);
}
return 0;
}