Codeforces 237E Build String 【费用流】
题目链接：Codeforces 237E Build String
E. Build String
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You desperately need to build some string t. For that you’ve got n more strings s1, s2, …, sn. To build string t, you are allowed to perform exactly |t| (|t| is the length of string t) operations on these strings. Each operation looks like that:
choose any non-empty string from strings s1, s2, …, sn;
choose an arbitrary character from the chosen string and write it on a piece of paper;
remove the chosen character from the chosen string.
Note that after you perform the described operation, the total number of characters in strings s1, s2, …, sn decreases by 1. We are assumed to build string t, if the characters, written on the piece of paper, in the order of performed operations form string t.
There are other limitations, though. For each string si you know number ai — the maximum number of characters you are allowed to delete from string si. You also know that each operation that results in deleting a character from string si, costs i rubles. That is, an operation on string s1 is the cheapest (it costs 1 ruble), and the operation on string sn is the most expensive one (it costs n rubles).
Your task is to count the minimum amount of money (in rubles) you will need to build string t by the given rules. Consider the cost of building string t to be the sum of prices of the operations you use.
Input
The first line of the input contains string t — the string that you need to build.
The second line contains a single integer n (1 ≤ n ≤ 100) — the number of strings to which you are allowed to apply the described operation. Each of the next n lines contains a string and an integer. The i-th line contains space-separated string si and integer ai (0 ≤ ai ≤ 100). Number ai represents the maximum number of characters that can be deleted from string si.
All strings in the input only consist of lowercase English letters. All strings are non-empty. The lengths of all strings do not exceed 100 characters.
Output
Print a single number — the minimum money (in rubles) you need in order to build string t. If there is no solution, print -1.
Examples
input
bbaze
3
bzb 2
aeb 3
ba 10
output
8
input
abacaba
4
aba 2
bcc 1
caa 2
bbb 5
output
18
input
xyz
4
axx 8
za 1
efg 4
t 1
output
-1
Note
Notes to the samples:
In the first sample from the first string you should take characters “b” and “z” with price 1 ruble, from the second string characters “a”, “e” и “b” with price 2 rubles. The price of the string t in this case is 2·1 + 3·2 = 8.
In the second sample from the first string you should take two characters “a” with price 1 ruble, from the second string character “c” with price 2 rubles, from the third string two characters “a” with price 3 rubles, from the fourth string two characters “b” with price 4 rubles. The price of the string t in this case is 2·1 + 1·2 + 2·3 + 2·4 = 18.
In the third sample the solution doesn’t exist because there is no character “y” in given strings.
题意：给定一个主串s，现在给你n个模式串，从第i个串中拿出一个字符的代价是i且最多拿出num[i]个字符。问你能否构造出s，若可以输出最小代价。
思路：很裸的费用流了。
我们从源到26个字符建边，容量为s串中该字符的个数，费用为0。
每个字符向n个模式串建边，记串i有cnt个该字符，那么就建一条容量为cnt，费用为i的边。
最后每个模式串向汇建边，容量为num[i]，费用为0。
最后结果满流说明可以构造出s串，反之不可以。
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <queue>
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e6 + 1;
const int MOD = 1073741824;
const int INF = 0x3f3f3f3f;
void add(LL &x, LL y) { x += y; x %= MOD; }
int num[40], use[110];
char str[110];
int cnt[110][40];
struct Edge {
int from, to, cap, flow, cost, next;
};
Edge edge[10000];
int head[200], edgenum;
void init() { CLR(head, -1); edgenum = 0; }
void addEdge(int u, int v, int w, int c) {
Edge E = {u, v, w, 0, c, head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
Edge E1 = {v, u, 0, 0, -c, head[v]};
edge[edgenum] = E1;
head[v] = edgenum++;
}
bool vis[200];
int dist[200], pre[200];
bool SPFA(int s, int t) {
queue<int> Q; CLR(dist, INF); CLR(vis, false); CLR(pre, -1);
vis[s] = true; dist[s] = 0; Q.push(s);
while(!Q.empty()) {
int u = Q.front(); Q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next) {
Edge E = edge[i];
if(dist[E.to] > dist[u] + E.cost && E.cap > E.flow) {
dist[E.to] = dist[u] + E.cost;
pre[E.to] = i;
if(!vis[E.to]) {
vis[E.to] = true;
Q.push(E.to);
}
}
}
}
return pre[t] != -1;
}
void MCMF(int s, int t, int &flow, int &cost) {
flow = cost = 0;
while(SPFA(s, t)) {
int Min = INF;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
Edge E = edge[i];
Min = min(Min, E.cap - E.flow);
}
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
edge[i].flow += Min;
edge[i^1].flow -= Min;
//cout << edge[i].cost << endl;
cost += edge[i].cost * Min;
}
flow += Min;
}
}
int main()
{
while(scanf("%s", str) != EOF) {
int len = strlen(str); CLR(num, 0);
int sum = len;
for(int i = 0; i < len; i++) {
int v = str[i] - 'a' + 1;
num[v]++;
}
int n; scanf("%d", &n); CLR(cnt, 0);
for(int i = 1; i <= n; i++) {
scanf("%s%d", str, &use[i]);
len = strlen(str);
for(int j = 0; j < len; j++) {
int v = str[j] - 'a' + 1;
cnt[i][v]++;
}
}
init();
int s = 0, t = 26+n+1;
for(int i = 1; i <= 26; i++) {
if(num[i] == 0) continue;
addEdge(s, i, num[i], 0);
for(int j = 1; j <= n; j++) {
if(cnt[j][i] == 0) continue;
addEdge(i, 26+j, cnt[j][i], j);
}
}
for(int i = 1; i <= n; i++) {
addEdge(26+i, t, use[i], 0);
}
int flow, cost;
MCMF(s, t, flow, cost);
if(flow != sum) {
printf("-1\n");
}
else {
printf("%d\n", cost);
}
}
return 0;
}