NOIP2012 Day2 T2(线段树/二分+前缀和+离线)

线段树做法：
裸模板，维护区间增加区间最小值，但是常数大容易T

二分+前缀和做法：
每次二分一个通过申请的个数，然后暴力前缀和判断，复杂度与线段树一样，但是常数小

线段树做法：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
using namespace std;
const int MAXN = 1000000 + 5;
int n, m, ri[MAXN];
#define lc (o << 1)
#define rc ((o << 1) | 1)
#define M ((l + r) >> 1)
int minv[MAXN * 4], addv[MAXN * 4];
void pushup(int o) {
    minv[o] = min(minv[lc], minv[rc]);
}
void pushdown(int o) {
    if (addv[o]) {
        addv[lc] += addv[o], addv[rc] += addv[o];
        minv[lc] += addv[o], minv[rc] += addv[o];
        addv[o] = 0;
    }
}
void build(int o, int l, int r) {
    if (l == r) minv[o] = ri[l]; else build(lc, l, M), build(rc, M + 1, r), pushup(o);
}
void update(int o, int l, int r, int x, int y, int v) {
    if (x <= l && r <= y) {
        minv[o] += v, addv[o] += v;
        return ;
    }
    pushdown(o);
    if (x <= M) update(lc, l, M, x, y, v);
    if (M < y) update(rc, M + 1, r, x, y, v);
    pushup(o);
}
int query(int o, int l, int r, int x, int y) {
    int ret = 2000000000;
    if (x <= l && r <= y) {
        return minv[o];
    }
    pushdown(o);
    if (x <= M) ret = min(ret, query(lc, l, M, x, y));
    if (M < y) ret = min(ret, query(rc, M + 1, r, x, y));
    pushup(o);
    return ret;
}
void clean() {
    ms(addv, 0);
}
void solve() {
    clean();
    for (int i = 1; i <= n; i++) scanf("%d", &ri[i]);
    build(1, 1, n);
    for (int i = 1; i <= m; i++) {
        int di, si, ti;
        scanf("%d%d%d", &di, &si, &ti);
        int ret = query(1, 1, n, si, ti);
        if (ret < di) {
            printf("-1\n%d\n", i);
            return ;
        }
        update(1, 1, n, si, ti, -di);
    }
    printf("0\n");
}
int main() {
    scanf("%d%d", &n, &m), solve();
    return 0;
}

二分+前缀和做法：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
using namespace std;
const int MAXN = 1000000 + 5;
int n, m, ri[MAXN];
int di[MAXN], si[MAXN], ti[MAXN];
int ci[MAXN];
bool check(int x) {
    for (int i = 0; i <= n + 1; i++) ci[i] = 0;
    for (int i = 1; i <= x; i++) ci[si[i]] += di[i], ci[ti[i] + 1] -= di[i];
    for (int i = 1; i <= n + 1; i++) {
        ci[i] += ci[i - 1];
        if (ci[i] > ri[i]) return false;
    }
    return true;
}
void clean() {}
void solve() {
    clean();
    for (int i = 1; i <= n; i++) scanf("%d", &ri[i]);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &di[i], &si[i], &ti[i]);
    }
    int ans = 0, l = 0, r = m + 1;
    while (l < r) {
        int mid = (l + r) / 2;
        if (check(mid)) l = mid + 1, ans = mid; else r = mid;
    }
    if (ans == m) printf("0\n"); else printf("-1\n%d\n", ans + 1);
}
int main() {
    scanf("%d%d", &n, &m), solve();
    return 0;
}