目录
- CSP201809-4 再卖菜
- 60分dfs
- 100分建立差分约束系统
CSP201809-4 再卖菜
这场打完就忘记了。当时这题一直在想怎么建立差分约束系统，最后没建出来0分，队友搜索暴力剪纸可能还有记忆化过的，其他人有的都是搜索拿的几十分，
用最暴力的写法直接搜即可。
60分dfs
菜的价格在1~100之间，直接dfs枚举菜的价格即可。300^n复杂度?其实减枝能剪掉很多。
#include
using namespace std;
const int MAXN = 310;
int n, a[MAXN], b[MAXN];
bool flag = false;
void dfs(int step) {
int i = 0;
if(flag) {
return ;
}
if(step == n + 1) {
bool ok = true;
if((b[1] + b[2]) / 2 != a[1] || (b[n] + b[n - 1]) / 2 != a[n]) {
ok = false;
}
for(i = 2; i <= n - 1; i++ ) {
if((b[i - 1] + b[i] + b[i + 1]) / 3 != a[i]) {
ok = false;
}
}
if(!ok) {
return ;
}
for(i = 1; i <= n; i++ ) {
if(i != 1) {
printf(" ");
}
printf("%d", b[i]);
}
puts("");
flag = true;
return ;
}
for(i = 1; i <= 130; i++ ) {
b[step] = i;
if(step == 2) {
if(a[1] != (b[1] + b[2]) / 2) {
continue;
}
} else {
if(step >= 3) {
if((b[step] + b[step - 1] + b[step - 2]) / 3 != a[step - 1]) {
continue;
}
}
}
if(step <= n) {
dfs(step + 1);
}
}
}
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; i++ ) {
scanf("%d", &a[i]);
}
dfs(1);
return 0;
} 100分建立差分约束系统
在现场的时候我就一直在想怎么建。三元不等式不会化简单。现在才知道，另si = x1 + x2 + ... xn
这样例如x2 + x3 + x4 就可以用s4 - s1代替。这样题目获得的是a数组。我们有s数组，不等式直接建立差分约束系统即可。具体不等式见注释
学过差分约束的讲到这里就够了，不会的需要补一补基础。
/**
等式的确立是由于向下取整
a1 * 2 <= x1 + x2 <= a1 * 2 + 1
a2 * 3 <= x1 + x2 + x3 <= a2 * 3 + 2
...
an * 2 <= x[n - 1] + x[n] <= an * 2 + 1
let
si = x0 + x1 + x2 + ... + xn
change xi
[1]
such as a1 * 2 <= s2 - s0
a1 * 2 + 1 <= s0 - s2
[2]
so
把上面x数组改s即可。
0 <==> 3
...
...
...
n - 3 <==> n
[3]
xi >= 1
for each xi - xj >= k add_edge(j -> i, w) find the longest path
*/
#include
using namespace std;
const int MAXN = 3050;
const int MAXM = MAXN * MAXN;
int n, m, first[MAXN], sign, a[MAXN];
struct Edge {
int to, w, next;
} edge[MAXM];
void init() {
memset(first, -1, sizeof(first));
sign = 0;
}
void add_edge(int u, int v, int w) {
edge[sign].to = v;
edge[sign].w = w;
edge[sign].next = first[u];
first[u] = sign++;
}
int dist[MAXN], inq[MAXN];
void spfa() {
queueque;
for(int i = 0; i <= n; i++ ) {
que.push(i);
dist[i] = 0, inq[i] = 0;
}
while(!que.empty()) {
int now = que.front();
que.pop();
inq[now] = 0;
for(int i = first[now]; ~i; i = edge[i].next) {
int to = edge[i].to, w = edge[i].w;
if(dist[to] < dist[now] + w) {
dist[to] = dist[now] + w;
if(!inq[to]) {
que.push(to);
inq[to] = 1;
}
}
}
}
}
int main()
{
scanf("%d", &n);
init();
for(int i = 1; i <= n; i++ ) {
scanf("%d", &a[i]);
}
for(int i = 0; i <= n - 3; i++ ) {
add_edge(i + 3, i, -(a[i + 2] * 3 + 2)); /// x[i] - x[i + 3] >= -(a[i+2] * 3 + 2)
add_edge(i, i + 3, a[i + 2] * 3); /// x[i +3] - x[i] >= (a[i+2] * 3)
}
add_edge(2, 0, -(a[1] * 2 + 1)); ///s0-s2 >= -(a[1] * 2 + 1)
add_edge(0, 2, a[1] * 2); ///s2-s0 >= (a[1] * 2)
add_edge(n, n - 2, -(a[n] * 2 + 1)); ///sn-s[n-2] >= -(a[n] * 2 + 1)
add_edge(n - 2, n, a[n] * 2); ///s[n-2]-sn >= (a[n] * 2)
for(int i = 1; i <= n; i++ ) { /// si - s[i-1] >= 1
add_edge(i - 1, i, 1);
}
spfa();
a[1] = dist[1];
for(int i = 2; i <= n; i++ ) {
a[i] = dist[i] - dist[i - 1];
}
for(int i = 1; i <= n; i++ ) {
if(i != 1) {
printf(" ");
}
printf("%d", a[i]);
}
puts("");
return 0;
}