★ Title Description
Only one of "o", "X" and a character string consisting of "*", the "*" replaced by "o" requires a cost, the cost of replacing the need b "X."
A string legitimate need to meet the following two conditions:
The number "o" of all prefix string must be not less than 1. The number of "x";
2. The number of the string "o" must be equal to the number of "x";
Hong Nima want to know all the "*" replace "o" or "x" so that the string is a legal minimum cost how much?
If no solution outputs -1.
★ input format
The first line a string S, the string represented;
Next m lines of two positive integers ai, bi, represents replaced by "o", the cost required for "x";
For 60% of the data string length is not greater than 1000;
To 100% of the data string length is not greater than 100000,0 <= ai, bi <= 108.
★ output format
If solutions, it outputs a positive integer representing the minimum cost;
If there is no solution, the output of -1.
★ sample input
o**x
1 2
3 5★ Sample Output
5★ Tips
no
★ reference code
/*
https://blog.csdn.net/f_zyj/article/details/74158391
优先队列优化的贪心问题。
优先队列 + 贪心
1.第一次遍历字符串，把“*”全部用“ x ”替换。在替换过程中记录一下所花代价 sum ，并把 a-b 值加入到一个队列；
2.开始第二次遍历字符串，此时记录 cnt，每当 cnt<0时，表明需要往前面字符串o字符不足，就可以选择队列中的最小的位置替换，等于“*”-> “x”-> “o”,此时 cnt+2、sum+(a-b)
*/
#include<bits/stdc++.h>
using namespace std;
int n;
char S[100005];
int A[100005]={0};
int B[100005]={0};
long long sum = 0; //记录代价
int cnt = 0; //记录o和x的个数差
priority_queue<int> pqi;
long long fun(){
for(int i=0; i<n; ++i){
if(S[i] == 'o') ++cnt;
else if(S[i]=='x' || S[i]=='*') --cnt;
if(S[i]=='*') pqi.push(B[i]-A[i]); //替换时，a[i] - b[i]，考虑优先出来大的，故取反
if(cnt < 0) {
if(pqi.empty()) return -1;
sum -= pqi.top();
pqi.pop();
cnt += 2;
}
}
if (cnt != 0) return -1;
return sum;
}
int main(){
scanf("%s", S);
n = (int)strlen(S);
for(int i=0; i<n; ++i){
if(S[i]=='*'){
scanf("%d%d",&A[i],&B[i]);
sum += B[i];
}
}
printf("%lld\n", fun());
return 0;
}