USACO 1.1 Broken Necklace (DP动态规划)

<pre name="code" class="cpp">#include <stdio.h>
#define DEBUG 0
#define TESTCASES 8

#define LEN_NECKLACE 350
#define MAX(a, b) ( (a) > (b) ? (a) : (b) )

int numOfBeads;
int numOfDoubleBeads;
char necklace[LEN_NECKLACE * 2 + 1];

// beadsCollectedForward[bead][0]表示当断点为bead-1和bead之间时, 从bead开始往前（右）连续收集到红色珠子的数量
// beadsCollectedForward[bead][1]表示当断点为bead-1和bead之间时, 从bead开始往前（右）连续收集到蓝色珠子的数量
int beadsCollectedForward[LEN_NECKLACE * 2 + 1][2];
// beadsCollectedBackward[bead][0]表示当断点为bead和bead+1之间时, 从bead开始往后（左）连续收集到红色珠子的数量
// beadsCollectedBackward[bead][1]表示当断点为bead和bead+1之间时, 从bead开始往后（左）连续收集到蓝色珠子的数量
int beadsCollectedBackward[LEN_NECKLACE * 2 + 1][2];

void strCpy(char *to, char *from){
while (*to = *from){
if (*from == '\0')
break;
to++;
from++;
}
*to = '\0';
}

void strCat(char *before, char *after){
while (*before != '\0')
before++;
while (*after != '\0'){
*before = *after;
before++;
after++;
}
*before = '\0';
}

void initialBeadsCollected(int beadsCollected[][2], int bead){
switch (necklace[bead]){
case 'w':
beadsCollected[bead][0] =  1;
beadsCollected[bead][1] =  1;
break;
case 'r':
beadsCollected[bead][0] = 1;
beadsCollected[bead][1] = 0;
break;
case 'b':
beadsCollected[bead][0] = 0;
beadsCollected[bead][1] = 1;
break;
}
}

void collect(int beadsCollected[][2], int bead, int direction){
switch (necklace[bead]){
case 'w':
beadsCollected[bead][0] = beadsCollected[bead - direction][0] + 1;
beadsCollected[bead][1] = beadsCollected[bead - direction][1] + 1;
break;
case 'r':
beadsCollected[bead][0] = beadsCollected[bead - direction][0] + 1;
beadsCollected[bead][1] = 0;
break;
case 'b':
beadsCollected[bead][0] = 0;
beadsCollected[bead][1] = beadsCollected[bead - direction][1] + 1;
break;
}
}

int main(){
#if DEBUG
int testCase;
for (testCase = 1; testCase <= TESTCASES; testCase++){
char inputFileName[20] = "necklace.inX";
inputFileName[11] = '1' +  (testCase - 1);
freopen(inputFileName, "r", stdin);
printf("#%d\n", testCase);
#endif

scanf("%d %s", &numOfBeads, necklace);

//两条链子相连当成环处理
numOfDoubleBeads = numOfBeads * 2;
char temp[LEN_NECKLACE * 2 + 1];
strCpy(temp, necklace);
strCat(necklace, temp);

int bead = 0;
initialBeadsCollected(beadsCollectedBackward, bead);
for (bead++; bead < numOfDoubleBeads; bead++)
collect(beadsCollectedBackward, bead, 1);
bead = numOfDoubleBeads - 1;
initialBeadsCollected(beadsCollectedForward, bead);
for (bead--; bead >= 0; bead--)
collect(beadsCollectedForward, bead, -1);

int maxBeadsCollected = 0;
int lastBead = numOfDoubleBeads - 1;
//注意边界，断点在bead和bead+1之间，从bead开始往后（左）收集，从bead+1开始往前（右）收集
for (bead = 0; bead < lastBead; bead++){
int beadsCollected = MAX(beadsCollectedBackward[bead][0], beadsCollectedBackward[bead ][1]) +
MAX(beadsCollectedForward[bead + 1][0], beadsCollectedForward[bead + 1][1]);
if (beadsCollected > maxBeadsCollected)
maxBeadsCollected = beadsCollected;
}

printf("%d\n", maxBeadsCollected > numOfBeads ? numOfBeads : maxBeadsCollected);

#if DEBUG
}
#endif
return 0;
}