Problem B: Passwords
题意：给出n个字符串密码，在给出一个正确密码，输密码必须先输入简单的密码，连续输错k次密码会罚时5秒，输一次密码耗时1秒，求可能的最短和最长的耗时。（注意：给出的n个密码可能包含多个正确密码）
思路：略
code：
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <iostream>
#include <algorithm>
#define rep(i,k,n) for(int i = k;i < n;i ++)
#define repp(i,k,n) for(int i = k;i <= n;i ++)
#define scan(d) scanf("%d",&d)
#define scanl(d) scanf("%I64d",&d)
#define scann(n,m) scanf("%d %d",&n,&m)
#define scannl(n,m) scanf("%I64d %I64d",&n,&m)
#define mst(a,k) memset(a,k,sizeof(a))
#define mod 1e9+7
#define ll long long
#define maxn 105
using namespace std;
int cnt[maxn];
map<string,int> mmap;
string p,s;
int main(void){
int n,k;
cin >> n >>k;
mst(cnt,0);
mmap.clear();
for(int i = 1;i <= n;i ++){
cin >> p;
mmap[p] ++;
cnt[p.size()] ++;
}
cin >> s;
int len = s.size();
int cnt1,cnt2;
cnt1 = cnt2 = 0;
for(int i = 1;i <= len - 1;i ++){
cnt1 += cnt[i];
}
cnt2 += cnt[len];
cnt2 -= (mmap[s]);
int ans1,ans2;
ans1 = (cnt1/k)*5 + 1 + cnt1;
ans2 = ((cnt1 + cnt2)/k)*5 + cnt1 + cnt2 + 1;
printf("%d %d\n",ans1,ans2);
return 0;
}
Problem C: Journey
题意：给出一个有n个节点的有向图，每条路包含一个权值表示走过这条路的时间，要求在给出的总时间T内必须从起点1走到终点n，求在满足条件的情况下，走过节点数的最大值。
思路：dfs + dp。dp[i][t]表示走到第i个节点路过节点数为t时使用的最小时间的，则dp[i][t] = min(dp[k][t-1]+t[k][i],dp[i][t])。
code:
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <iostream>
#include <algorithm>
#define rep(i,k,n) for(int i = k;i < n;i ++)
#define repp(i,k,n) for(int i = k;i <= n;i ++)
#define scan(d) scanf("%d",&d)
#define scanl(d) scanf("%I64d",&d)
#define scann(n,m) scanf("%d %d",&n,&m)
#define scannl(n,m) scanf("%I64d %I64d",&n,&m)
#define mst(a,k) memset(a,k,sizeof(a))
#define mod 1e9+7
#define inf 2000000005
#define ll long long
#define maxn 5005
using namespace std;
int edgesCnt;
int head[maxn];
int fa[maxn][maxn]; //fa[i][j]表示走到第i个节点时走过了j个节点时的前驱
int dp[maxn][maxn]; //dp[i][t]表示走到第i个节点路过节点数为t时使用的最小时间的，则dp[i][t] = min(dp[k][t-1]+t[k][i],dp[i][t])。
int n,m,Tot;
struct Edge{
int to;
int dist;
int next;
}edges[maxn];
void init(){
edgesCnt = 0;
memset(head,-1,sizeof(head));
}
void addEdge(int from,int to,int d){
edges[edgesCnt].dist = d;
edges[edgesCnt].to = to;
edges[edgesCnt].next = head[from];
head[from] = edgesCnt;
edgesCnt ++;
}
void dfs(int u,int step){
for(int j = head[u];j != -1;j = edges[j].next){
int v = edges[j].to;
int d = edges[j].dist;
if(dp[v][step+1] > dp[u][step]+d && dp[u][step]+d <= Tot){
dp[v][step+1] = dp[u][step]+d;
fa[v][step+1] = u;
dfs(v,step+1);
}
}
}
//递归输出路径
void print(int u,int j){
if(u == 1) {
printf("1 ");
return;
}
else{
print(fa[u][j],j-1);
}
printf("%d ",u);
if(u == n)
printf("\n");
}
int main(void){
cin >> n >> m >> Tot;
init();
for(int i = 1;i <= m;i ++){
int u,v,d;
scanf("%d %d %d",&u,&v,&d);
addEdge(u,v,d);
}
for(int i = 1;i <= n+1;i ++){
for(int j = 1;j <= n+1;j ++){
dp[i][j] = inf;
}
}
dp[1][1] = 0;
dfs(1,1);
for(int i = n;i >= 1;i --){
if(dp[n][i] <= Tot){
printf("%d\n",i);
print(n,i);
return 0;
}
}
return 0;
}
Problem D: Maxim and Array
题意：给出一个包含n个数的数组，给出一个整数x，执行k次操作（对数组任意一个元素加x或减x）。求数组的积最小。
思路：若数组的积的符号是负，则每次操作只需要对绝对值最小的元素执行操作；若数组的积是正，则同样是最绝对值最小的元素进行操作，目的是尽快实现变号，若无法变号，也可以使积变小。综上，只需使用优先队列处理至多k次操作即可。
code:
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <iostream>
#include <algorithm>
#define rep(i,k,n) for(int i = k;i < n;i ++)
#define repp(i,k,n) for(int i = k;i <= n;i ++)
#define scan(d) scanf("%d",&d)
#define scanl(d) scanf("%I64d",&d)
#define scann(n,m) scanf("%d %d",&n,&m)
#define scannl(n,m) scanf("%I64d %I64d",&n,&m)
#define mst(a,k) memset(a,k,sizeof(a))
#define mod 1e9+7
#define ll long long
#define maxn 200005
using namespace std;
ll a[maxn];
ll n,k,x;
int flag;
struct node{
int id;
ll value;
bool flag; //状态，1代表正，0代表负
bool operator < (const node& rhs) const{
return value > rhs.value;
}
};
int main(void){
while(cin >> n >> k >> x){
priority_queue<node> Q;
//Q.clear();
while(!Q.empty()) Q.pop();
int cnt = 0;
for(int i = 1;i <= n;i ++) {
cin >> a[i];
node p;
if(a[i] < 0) {
cnt ++;
p.value = abs(a[i]);
p.flag = 0;
p.id = i;
Q.push(p);
}
else{
p.value = abs(a[i]);
p.flag = 1;
p.id = i;
Q.push(p);
}
}
flag = cnt%2?-1:1;
ll sum = x*k;
while(k){
node tp = Q.top();
Q.pop();
ll id,value,f;
id = tp.id;
value = tp.value;
f = tp.flag;
if(flag == 1){ //符号为正
if(sum > value){
a[id] -= f == 1?(value/x)*x:(-value/x)*x;
k -= (value/x);
sum = k*x;
if(a[id] >= 0 && f == 1) {a[id] -= x; k --; sum -= x;}
if(a[id] < 0 && f == 0) {a[id] += x; k --; sum -= x;}
flag = -1;
node p;
p.flag = f == 1?0:1;
p.id = id;
p.value = abs(a[id]);
Q.push(p);
}
else{
a[id] -= f == 1?sum:-sum;
k = sum = 0;
}
}
else{ //符号为负
a[id] += f == 1?x:-x;
k --;
sum -= x;
node p;
p.flag = f == 1?1:0;
p.id = id;
p.value = abs(a[id]);
Q.push(p);
}
}
for(int i = 1;i <= n;i ++) printf("%I64d ",a[i]);
cout << endl;
}
return 0;
}
时间： 2024-10-12 02:47:06