UVALive 6042 Bee Tower

2014-11-24 01:32:52 · 作者: · 浏览: 2
dp。只需要对于每个最高点,从左边递推一遍,再从右边递推一遍。找最小的花费就可以了。
#include  
#include  
#include  
#include  
#include  
#include  
#define LL long long  
#define CLR(a, b) memset(a, b, sizeof(a))  
  
using namespace std;  
const int N = 111;  
const int W = 555;  
const int INF = 0X3f3f3f3f;  
  
struct Power  
{  
    int p, h;  
} pw[N];  
  
int dp[N][W], h, w;  
  
int solvel(int s, int e)  
{  
    int ret = INF;  
    if(pw[s].h <= h) return 0;  
    dp[s][pw[s].p] = 0;  
    for(int i = s - 1; i >= e; i --)  
    {  
        if(pw[i + 1].h - pw[i].h > h) return INF;  
        for(int j = pw[s].p - (s - i - 1); j >= pw[s].p - (s - i - 1) * w; j --)  
        {  
            if(j > 0)for(int k = j - 1; k >= j - w; k --)  
            {  
                dp[i][k] = min(dp[i][k], dp[i + 1][j] + abs(pw[i].p - k) * pw[i].h);  
                if(pw[i].h <= h) ret = min(ret, dp[i][k]);  
            }  
        }  
    }  
    return ret;  
}  
  
int solver(int s, int e)  
{  
    int ret = INF;  
    dp[s][pw[s].p] = 0;  
    if(pw[s].h <= h) return 0;  
    for(int i = s + 1; i <= e; i ++)  
    {  
        if(pw[i - 1].h - pw[i].h > h) return INF;  
        for(int j = pw[s].p + (i - s - 1); j <= pw[s].p + (i - s - 1) * w; j ++)  
        {  
            if(j >
0)for(int k = j + 1; k <= j + w; k ++) { dp[i][k] = min(dp[i][k], dp[i - 1][j] + abs(pw[i].p - k) * pw[i].h); if(pw[i].h <= h) ret = min(ret, dp[i][k]); } } if(pw[i].h <= h) break; } return ret; } vector ca; int main() { int t, n, sit, hh, ans, cas = 1; scanf("%d", &t); while(t --) { scanf("%d%d%d", &n, &h, &w); hh = 0; ca.clear(); for(int i = 1; i <= n; i ++) { scanf("%d%d", &pw[i].p, &pw[i].h); if(pw[i].h > hh) { hh = pw[i].h; ca.clear(); ca.push_back(i); } else if(pw[i].h == hh) ca.push_back(i); } CLR(dp, INF); ans = INF; for(int i = 0; i < ca.size(); i ++) { ans = min(ans, solvel(ca[i], 1)); ans = min(ans, solver(ca[i], n)); } if(ans == INF) ans = -1; printf("Case #%d: %d\n", cas ++, ans); } }