uva 1400 "Ray, Pass me the dishes!"

这题属于数据结构题，用线段树做，不多说，看代码，结构体变量名有注释。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 500005
#define mem(a) memset(a, 0, sizeof(a))
using namespace std;

struct node
{
    long long vl;//区间左最大值
	long long vr;//区间右最大值
	long long w;//区间最大值
	long long sum;//区间和
    int l;//区间结果左坐标
	int r;//区间结果右坐标
	int yl;//区间左最大值的右坐标
	int xr;//区间右最大值的左坐标
}e[maxn * 4];

long long v[maxn];

void build(int l, int r, int n)
{
    if(l == r)
    {
        e
.sum = e
.w = e
.vl = e
.vr = v[l];
        e
.l = e
.r = e
.yl = e
.xr = l;
        return ;
    }
    int mid = (l + r) / 2;
    build(l, mid, 2 * n);
    build(mid + 1, r, 2 * n + 1);
    e
.sum = e[2 * n].sum + e[2 * n + 1].sum;
    long long wl = e[2 * n].w, wr = e[2 * n + 1].w, vm = e[2 * n].vr + e[2 * n + 1].vl;
    if(wl >= wr&&wl >= vm)
	{
	    e
.w = wl;
		e
.l = e[2 * n].l;
		e
.r = e[2 * n].r;
	}
    else if(vm >= wr)
	{
	    e
.w = vm;
		e
.l = e[2 * n].xr;
		e
.r = e[2 * n + 1].yl;
	}
    else 
	{
	    e
.w = wr;
		e
.l = e[2 * n + 1].l;
		e
.r = e[2 * n + 1].r;
	}
    if(e[2 * n].vl >= e[2 * n].sum + e[2 * n + 1].vl)
    {
        e
.vl = e[2 * n].vl;
        e
.yl = e[2 * n].yl;
    }
    else
    {
        e
.vl = e[2 * n].sum + e[2 * n + 1].vl;
        e
.yl = e[2 * n + 1].yl;
    }
    if(e[2 * n + 1].vr > e[2 * n].vr + e[2 * n + 1].sum)
    {
        e
.vr = e[2 * n + 1].vr;
        e
.xr = e[2 * n + 1].xr;
    }
    else
    {
        e
.vr = e[2 * n + 1].sum + e[2 * n].vr;
        e
.xr = e[2 * n].xr;
    }
}

node res(int n, int l, int r, int ll, int rr)
{
    if(ll == l&&rr == r)
	return e
;
    int mid = (rr + ll) / 2;
    if(r <= mid)
	return res(2 * n, l, r, ll, mid);
    else if(l > mid)
	return res(2 * n + 1, l, r, mid + 1, rr);
    else
    {
        node sonl, sonr, rst;
        sonl = res(2 * n, l, mid, ll, mid);
        sonr = res(2 * n + 1, mid + 1, r, mid + 1, rr);
        rst.sum = sonl.sum + sonr.sum;
        if(sonl.w >= sonr.w&&sonl.w >= sonl.vr + sonr.vl)
		{
		    rst.w = sonl.w;
			rst.l = sonl.l;
			rst.r = sonl.r;
		}
        else if(sonl.vr + sonr.vl >= sonr.w)
		{
		    rst.w = sonl.vr + sonr.vl;
			rst.l = sonl.xr;
			rst.r = sonr.yl;
		}
        else 
		{
		    rst.w = sonr.w;
			rst.l = sonr.l;
			rst.r = sonr.r;
		}
        if(sonl.vl >= sonl.sum + sonr.vl)
        {
            rst.vl = sonl.vl;
            rst.yl = sonl.yl;
        }
        else
        {
            rst.vl = sonl.sum + sonr.vl;
            rst.yl = sonr.yl;
        }
        if(sonr.vr > sonl.vr + sonr.sum)
        {
            rst.vr = sonr.vr;
            rst.xr = sonr.xr;
        }
        else
        {
            rst.vr = sonr.sum + sonl.vr;
            rst.xr = sonl.xr;
        }
        return rst;
    }
}

int main()
{
    int num, m, o = 0, i, j, k, x, y, l, r;
    while(scanf("%d%d",&num, &m) != EOF)
    {
    	mem(v);
    	mem(e);
        for(i = 1;i <= num;i++)
        {
            scanf("%lld",&v[i]);
        }
        build(1, num, 1);
        node ans;
        printf("Case %d:\n",++o);
        for(i = 0;i < m;i++)
        {
            scanf("%d%d", &l, &r);
            ans = res(1, l, r, 1, num);
            printf("%d %d\n", ans.l, ans.r);
        }
    }
    return 0;
}