版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/GYH0730/article/details/82459743
题目链接:https://www.luogu.org/recordnew/show/10501734
【问题分析】
求一个状态到另一个状态变换的最少费用,最短路径问题。
【建模方法】
软件的状态用二进制位表示,第i位为第i个错误是否存在。把每个状态看做一个顶点,一个状态应用一个补丁到达另一状态,连接一条权值为补丁时间的有向边。求从初始状态到目标状态的最短路径即可。
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int B1[105],B2[105],F1[105],F2[105];
int n,m;
char str1[105],str2[105];
int cost[105];
bool vis[10000005];
struct node
{
int state,cost;
node(){}
node(int _state,int _cost)
{
state = _state;
cost = _cost;
}
bool operator < (const struct node& a) const
{
return cost > a.cost;
}
};
int ans;
void BFS()
{
priority_queue<node> pq;
pq.push(node(((1 << n) - 1),0));
struct node t;
int flag;
while(!pq.empty()) {
t = pq.top();
pq.pop();
if(vis[t.state]) continue;
vis[t.state] = true;
for(int i = 1; i <= m; i++) {
flag = 0;
for(int j = 1; j <= n; j++) {
if( (B1[i] & (1 << (j - 1))) && !(t.state & (1 << (j - 1))) ) {
flag = 1;
break;
}
}
if(flag) continue;
flag = 0;
for(int j = 1; j <= n; j++) {
if( (B2[i] & (1 << (j - 1))) && (t.state & (1 << (j - 1))) ) {
flag = 1;
break;
}
}
if(flag) continue;
int _state = t.state;
flag = 0;
for(int j = 1; j <= n; j++) {
if((F1[i] & (1 << (j - 1))) && (t.state & (1 << (j - 1)))) {
_state ^= (1 << (j - 1));
}
}
for(int j = 1; j <= n; j++) {
if((F2[i] & (1 << (j - 1))) && !(t.state & (1 << (j - 1)))) {
_state |= (1 << (j - 1));
}
}
if(!_state) {
ans = min(ans,t.cost + cost[i]);
}
pq.push(node(_state,t.cost + cost[i]));
}
}
}
int main(void)
{
scanf("%d %d",&n,&m);
for(int i = 1; i <= m; i++) {
scanf("%d",&cost[i]);
scanf("%s %s",str1,str2);
for(int j = 0; j < n; j++) {
if(str1[j] == '+') B1[i] |= (1 << (n - j - 1));
if(str1[j] == '-') B2[i] |= (1 << (n - j - 1));
if(str2[j] == '-') F1[i] |= (1 << (n - j - 1));
if(str2[j] == '+') F2[i] |= (1 << (n - j - 1));
}
}
ans = INF;
BFS();
printf("%d\n",ans == INF ? 0 : ans);
return 0;
}