初见安~这里是传送门：洛谷P2762 太空飞行计划
【网络流|洛谷·[网络流24题]太空飞行计划问题】![网络流|洛谷·[网络流24题]太空飞行计划问题](https://img.it610.com/image/info8/e9ac8a2f64534448bbb420856dbc58a9.jpg)
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Sol
这个题老鬼畜了……其实就是读入格式很恶心。
具体做法其实跑一个最大流就好了，明显是一个匹配问题。
直接贴代码吧【懒……但也是真的水。
#include
#include
#include
#include
#include
#include
#define maxn 500
#define maxm 200005
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
bool flag;
char ch;
int read() {
int x = 0, f = 1, ch = getchar();
while(!isdigit(ch)) {if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
return x * f;
}struct edge {int to, w, nxt;
} e[maxm];
int head[maxn], k = 0;
void add(int u, int v, int w) {
e[k] = {v, w, head[u]};
head[u] = k++;
e[k] = {u, 0, head[v]};
head[v] = k++;
}int d[maxn], S, T;
queue q;
bool bfs() {
memset(d, 0, sizeof d);
while(q.size()) q.pop();
q.push(S);
d[S] = 1;
while(q.size()) {
register int u = q.front(), v;
q.pop();
for(int i = head[u];
~i;
i = e[i].nxt) {
v = e[i].to;
if(e[i].w && !d[v]) {
q.push(v), d[v] = d[u] + 1;
if(v == T) return true;
}
}
}
return false;
}int Dinic(int u, int flow) {
if(u == T) return flow;
register int rest = flow, k;
for(int i = head[u], v;
~i && rest;
i = e[i].nxt) {
v = e[i].to;
if(e[i].w && d[v] == d[u] + 1) {
k = Dinic(v, min(rest, e[i].w));
if(!k) d[v] = 0;
e[i].w -= k, e[i ^ 1].w += k;
rest -= k;
}
}
return flow - rest;
}int n, m, sum = 0;
signed main() {
memset(head, -1, sizeof head);
n = read(), m = read();
S = 0, T = n + m + 1;
for(int i = 1, x;
i <= n;
i++) {//oh，这恶心的读入……而且你还需要看懂在干什么
x = read();
sum += x;
add(S, i, x);
//连边，只给x元保证不会亏
char tools[10000];
memset(tools,0,sizeof tools);
cin.getline(tools,10000);
int ulen=0,tool;
while (sscanf(tools+ulen,"%d",&tool)==1) {
add(i, tool + n, INF);
//连边，匹配
if (tool==0) ulen++;
else while (tool) {
tool/=10;
ulen++;
}
ulen++;
}
}
for(int i = 1, x;
i <= m;
i++) x = read(), add(i + n, T, x);
//连边，仪器开销。
register int flow, ans = 0;
while(bfs()) while(flow = Dinic(S, INF)) ans += flow;
//最大流保证每个实验的仪器开销一定满流，除非会亏
for(int i = 1;
i <= n;
i++) if(d[i]) printf("%d ", i);
puts("");
for(int i = n + 1;
i <= n + m;
i++) if(d[i]) printf("%d ", i - n);
//找匹配
printf("\n%d\n", sum - ans);
return 0;
} 咕咕咕了好久的一篇文章……
迎评：）
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