题目传送
dp是常规的;(m^2)的预处理:把位置存进vector然后(O(1))算出想要的;WA点:要注意特意设置一下val[i][v.size()]=0
,即全天都放鸽子则花费时间为0.
#include <bits/stdc++.h>
using namespace std;
int T, n, m, d, ans;
int val[205][205], dp[205][205];
char s[205];
int main() {
for (scanf("%d", &T); T; T--) {
memset(val, 0x3f, sizeof val);
memset(dp, 0x3f, sizeof dp);
scanf("%d %d %d", &n, &m, &d);
for (int i = 1; i <= n; i++) {
scanf("%s", s + 1);
vector<int> v;
for (int j = 1; j <= m; j++) {
if (s[j] == '1') {
v.push_back(j);
}
}
val[i][v.size()] = 0;
for (int j = 0; j < v.size(); j++) {
for (int k = j; k < v.size(); k++) {
int l = v[j], r = v[k];
int angry = v.size() - (k - j + 1);
val[i][angry] = min(val[i][angry], r - l + 1);
}
}
}
for (int i = 0; i <= d; i++)
dp[0][i] = 0;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= d; j++) {
for (int k = 0; k <= j; k++)
if (val[i][k] < 0x3f3f3f3f && dp[i - 1][j - k] < 0x3f3f3f3f)
dp[i][j] = min(dp[i][j], dp[i - 1][j - k] + val[i][k]);
}
}
ans = 0x3f3f3f3f;
for (int i = 0; i <= d; i++)
ans = min(ans, dp[n][i]);
printf("%d
", ans);
}
}