BZOJ 4276 费用流+线段树构图

来源：互联网发布：淘宝丝袜真人秀编辑：程序博客网时间：2024/05/05 08:26

为什么别人22s我37s。。

发现原来是写的struct new

改成namespace变成31s了

又一道线段树构图

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 205005, M = 500005;const int inf = 0x3f3f3f3f;#define FOR(i,j,k) for(i=j;i<=k;i++)int read() {    int s = 0, f = 1; char ch = getchar();    for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;    for (; '0' <= ch && ch <= '9'; ch = getchar()) s = s * 10 + ch - '0';    return s * f;}namespace CostFlow {    int h[N], p[M], v[M], w[M], c[M], vis[N], d[N], pre[N], q[N * 5], cnt;    int s, t, ans;    void init(int _s, int _t) {        s = _s; t = _t; cnt = 0;        memset(h, -1, sizeof h);    }    void add(int i, int j, int k, int l) {        p[cnt] = h[i]; v[cnt] = j; w[cnt] = k; c[cnt] = l; h[i] = cnt ++;        p[cnt] = h[j]; v[cnt] = i; w[cnt] = 0; c[cnt] = -l; h[j] = cnt ++;    }    bool spfa() {        memset(d, 0xef, sizeof d);        memset(vis, 0, sizeof vis);        int f = 1, r = 1, u, i;        q[r++] = s; d[s] = 0; pre[s] = pre[t] = -1; vis[s] = 1;        while (f < r) {            u = q[f++];            for (i = h[u]; i != -1; i = p[i])                if (w[i] > 0 && d[v[i]] < d[u] + c[i]) {                    d[v[i]] = d[u] + c[i];                    pre[v[i]] = i ^ 1;                    if (!vis[v[i]]) {                        q[r++] = v[i];                        vis[v[i]] = 1;                    }                }            vis[u] = 0;        }        return d[t] > 0;    }    void end() {        int u, sum = inf;        for (u = pre[t]; u != -1; u = pre[v[u]])            sum = min(sum, w[u ^ 1]);        for (u = pre[t]; u != -1; u = pre[v[u]])            w[u] += sum, w[u ^ 1] -= sum, ans += sum * c[u ^ 1];    }    int solve() {        ans = 0;        while (spfa()) end();        return ans;    }}void build(int k, int l, int r) {    if (l == r) {        CostFlow::add(k, CostFlow::t, 1, 0);        return;    }    int mid = l + r >> 1;    build(k * 2, l, mid); build(k * 2 + 1, mid + 1, r);    CostFlow::add(k, k * 2, inf, 0);    CostFlow::add(k, k * 2 + 1, inf, 0);}void query(int k, int l, int r, int ql, int qr, int x) {    if (ql == l && r == qr) {        CostFlow::add(x, k, inf, 0);        return;    }    int mid = l + r >> 1;    if (qr <= mid) query(k * 2, l, mid, ql, qr, x);    else if (mid < ql) query(k * 2 + 1, mid + 1, r, ql, qr, x);    else query(k * 2, l, mid, ql, mid, x), query(k * 2 + 1, mid + 1, r, mid + 1, qr, x);}int main() {    static int a[N], b[N], c[N];    int i, n = read(), rangel = 10000, ranger = 0;    CostFlow::init(0, N - 1);    FOR(i,1,n) {        a[i] = read(); b[i] = read(); c[i] = read();        rangel = min(rangel, a[i]);        ranger = max(ranger, --b[i]);    }    build(1, rangel, ranger);    FOR(i,1,n) {        int id = 30000 + i;        query(1, rangel, ranger, a[i], b[i], id);        CostFlow::add(CostFlow::s, id, 1, c[i]);    }    printf("%d", CostFlow::solve());    return 0;}

4276: [ONTAK2015]Bajtman i Okrągły Robin

Time Limit: 40 Sec Memory Limit: 256 MB
Submit: 132 Solved: 73
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Description

有n个强盗，其中第i个强盗会在[a[i],a[i]+1]，[a[i]+1,a[i]+2]，...，[b[i]-1,b[i]]这么多段长度为1时间中选出一个时间进行抢劫，并计划抢走c[i]元。作为保安，你在每一段长度为1的时间内最多只能制止一个强盗，那么你最多可以挽回多少损失呢？

Input

第一行包含一个正整数n(1<=n<=5000)，表示强盗的个数。

接下来n行，每行包含三个正整数a[i],b[i],c[i](1<=a[i]<b[i]<=5000,1<=c[i]<=10000)，依次描述每一个强盗。

Output

输出一个整数，即可以挽回的损失的最大值。

Sample Input

4
1 4 40
2 4 10
2 3 30
1 3 20

Sample Output

0 0