洛谷P2792 [JSOI2008]小店购物(最小树形图)

题意

题目链接

Sol

一开始的思路：新建一个虚点向每个点连边，再加上题面中给出的边，边权均为大小*需要购买的数量

然后发现死活都过不去

看了题解才发现题目中有个细节——买了\(A\)就可以买\(B\)，但是人家没告诉你必须买够\(A\)的数量才能买\(B\)呀qwqqqqqqq

所以建图的时候只算一次贡献就行了

这样剩下的个数都可以通过最小值买到

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c -'0', c = getchar();
    return x * f;
}
int N, M, R, m[MAXN], id[MAXN], vis[MAXN], fa[MAXN];
double mn[MAXN], val[MAXN];
struct Edge {
    int u, v; double w; int nxt;
}E[MAXN];
int head[MAXN], num = 1;
inline void AddEdge(int x, int y, double w) {
    E[num] = (Edge) {x, y, w, head[x]}; head[x] = num++;
}
double ZhuLiu() {
    double ans = 0; R = N;
    while("Liang Liang") {
        for(int i = 1; i <= N; i++) id[i] = vis[i] = 0, mn[i] = 1e9; int cnt = 0, x;
        for(int i = 1; i <= num; i++)
            if(E[i].v != E[i].u && (mn[E[i].v] > E[i].w))
                mn[E[i].v] = E[i].w, fa[E[i].v] = E[i].u;
        mn[R] = 0;
        for(int i = 1; i <= N; i++) {
            ans += mn[i];
            for(x = i; (!id[x]) && (vis[x] != i) && x != R; x = fa[x]) vis[x] = i; //tag
            if(x != R && (!id[x])) {
                id[x] = ++cnt;
                for(int t = fa[x]; t != x; t = fa[t]) id[t] = cnt;
            }
        }
        if(cnt == 0) return ans;
        for(int i = 1; i <= N; i++) if(!id[i]) id[i] = ++cnt;
        for(int i = 1; i <= num; i++) {
            double pre = mn[E[i].v];
            if((E[i].u = id[E[i].u]) != (E[i].v = id[E[i].v])) E[i].w -= pre;
        }
        N = cnt; R = id[R];
    }
    return ans;
}
int main() {
    N = read();
    for(int i = 1; i <= N; i++) {
        scanf("%lf", &val[i]), m[i] = read(), AddEdge(N + 1, i, val[i]);
    }
    N++;
    M = read();
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read(); double z; scanf("%lf", &z);
        AddEdge(x, y, z); val[y] = min(val[y], z);
    }
    double ans = 0;
    for(int i = 1; i <= N - 1; i++) ans += 1.0 * (m[i] - 1) * val[i];// printf("%d\n", m[i] - 1);
    printf("%.2lf", ans + ZhuLiu());
    return 0;
}